IN   MEMORIAM 
FLORIAN  CAJORl 


"^(^0^        ^t*Y^. 


INDUCTIVE 


PLANE    GEOMETEY 


WITE 

NUMEB0U8  EXEBCISES,    THEOREMS,  AND  PB0BLEM8 
FOB  ADVANCE   WOBK 


BY 


G.   IRVING  HOPKINS 
•I 

INSTRUCTOR  IN    MATHEMATICS  AND  ASTRONOMY 
HIGH  SCHOOL,   MANCHBSTKR,    N.H. 


'ji'mrsi^D  jf^hiTi'ojfr 


BOSTON,   U.S.A. 

D.  C.  HEATH  &  CO.,  PUBLISHERS 

1902 


COPTEIGHT,  1891  AND  1902, 

By  G.  I.  HOPKINS. 


CAJORl 


k 


PREFACE. 

The  inductive  method  of  teaching  geometry  is  recognized 
as  the  ideal  one  by  all  progressive  teachers.  Committing  to 
memory  the  demonstration  of  another  is  acknowledged  to  be 
of  little  benefit  to  the  pupil.  Development  of  the  reasoning 
powers  comes  only  from  their  judicious  exercise ;  and  this  is 
possible  only  in  small  degree  from  merely  yielding  assent  to 
the  logic  of  another. 

In  an  experience  of  twenty  years  the  author  has  found  that 
fully  three  fourths  of  his  pupils  can  demonstrate  unaided,  or 
at  most  with  a  suggestion  or  two,  the  majority  of  the  theorems, 
the  demonstrations  of  which  are  given  in  most  text-books  for 
the  pupil  to  read  and  memorize.  Consequently  he  has  en- 
deavored to  be  consistent,  and  offer  aid  in  the  way  of  sugges- 
tions only  where  the  pupil  needs  it.  In  most  text-books  the 
proof  of  the  following  easy  theorem  is  given  for  the  pupil  to 
memorize ;  viz. :  "  If  two  parallel  lines  are  cut  by  a  third 
straight  line  the  exterior  interior  angles  are  equal."  On  the 
very  same  page  are  given  others,  far  more  difficult,  for  the 
pupil  to  prove  unaided.  The  same  condition  of  things  is 
found  on  many  subsequent  pages.  The  author's  experience 
has  been  that  nine  tenths  of  his  pupils  demonstrate  the  above 
theorem  correctly  unaided,  and  that  the  other  tenth  need  only 
a  single  suggestion.  The  same  is  true  of  a  large  number  of 
the  theorems  and  construction  problems  usually  given  in  full. 

In  this  revised  edition,  the  arrangement  of  the  sequence  of 
theorems  has  been  radically  changed  in  many  places,  noticeably 
in  placing  the  subject  of  triangles  as  near  the  beginning  as 
possible.     This  is  done  because  of  the  great  advantage  in  the 

918232 


iy  Preface. 

use  of  equality  of  triangles  in  subsequent  demonstrations. 
This  aids  in  reducing  to  a  minimum  the  method  of  superposi- 
tion which  usually  confuses  the  beginner. 

The  author  has  endeavored  to  remedy  the  defects  of  the  old 
edition,  and  in  this  has  been  materially  aided  by  various  sug- 
gestions from  numerous  teachers  who  are  in  sympathy  with 
the  plan,  and  to  whom  he  takes  this  occasion  to  express  his 
thanks.  Thanks  are  also  due  the  publishers  for  the  excellence 
of  the  mechanical  work,  as  well  as  for  many  suggestions  as  to 
arrangement.  . 

finally,  one  of  the  most  important  features,  m  the  estima- 
tion of  the  author,  and  one  which  he  hopes  will  commend 
itself  to  teachers,  is  the  printing  the  essential  theorems  m  dif- 
ferent type  from  the  rest,  so  that  a  glance  at  the  page  will 
disclose  them ;  e.g. : 

If  two  sides  and  included  angle  of  one  triangle  are  equal  respectively 
to  two  sides  and  included  angle  of  another,  the  two  triangles  are  equal. 
Many  of  the  others  are  helpful  in  proving  subsequent  theo- 
rems, while  all  are  useful  as  exercises. 

G.  I.  H. 

Manchester,  N.H., 
June,  1902. 


CONTENTS. 


PAGE 

Definitions 1 

General  Axioms ,        .  7 

Particular  Axioms 8 

Symbols 12 

Abbreviations 12 

Theorems 14 

Triangles 16 

Transversals 21 

Angle  Measurement 32 

Advance  Theorems 38,  68,  97 

Quadrilaterals 35 

Circles 42 

Ratio  and  Proportion 63 

Theory  of  Limits 61 

Proportional  Lines 70 

Similar  Figures 74 

Polygons 78 

Problems  of  Computation 82,  100,  121 

Projection 84 

Areas 84 

V 


k 


^j  Contents. 


PAGE 

105 


Measurement  of  the  Circle 

126 
Problems  op  Construction 

Miscellaneous  Plane  Problems  for  Advance  Work  : 

r^  .        1  ...      147 

I.     Triangles 

II.     Quadrilaterals 

151 

III.  Circles 

IV.  Transformation  of  Figures  .        •        .      .  •        •        • 

154 
V.     Division  of  Figures 

159 
Solutions  by  Algebra 

...     164 
Special  Theorems 

...     176 
Maxima  and  Minima 

.     185 
Symmetry 

.     187 
Theorems  on  Symmetry 

189 
Tables  op  Metric  and  Common  Measures 

...    191 
Examination  Questions 


PLANE   GEOMETRY. 


3:^c 


INTRODUCTION. 

1.  Space  is  indefinite  extension  in  every  direction. 

2.  A  Material  Substance  is  anything,  large  or  small,  solid, 
liquid,  or  aeriform,  visible  or  invisible,  that  occupies  space. 

It  therefore  follows  that  material  substances  have  limited 
extension  in  every  direction. 

3.  For  purposes  of  measurement,  extension  in  three  direc- 
tions only  are  considered,  called,  respectively,  lengthy  breadth, 
and  thickness;  they  are  also  called,  collectively,  dimensions. 

4.  Magnitude,  in  general,  means  size,  and  is  applied  to  any- 
thing of  which  greater  or  less  can  be  predicated,  as  time,  weight, 
distance,  etc. 

A  Geometrical  Magnitude  is  that  which  has  one  or  more  of 
the  three  dimensions,  as  lines,  angles,  etc. 

5.  A  Geometrical  Point  has  position  merely;  i.e.  it  has  no 
magnitude. 

The  dots  made  by  pencil  and  crayon  are  called  points,  but 
they  are  really  small  substances  used  to  indicate  to  the  eye 
the  location  of  the  geometrical  point. 

6.  A  Geometrical  Line  has  only  one  dimension;  i.e.  length. 
The  lines  made  by  pencil  and  crayon  are  substances,  and 

may  be  called  physical  lines,  which  serve  to  show  the  position 
of  the  geometrical  lines. 

1 


2  Plane  Geometry. 

7.  A  Straight  Line  is  one  that  lies  evenly  between  its 
extreme  points. 

This  is  the  definition  as  given  by  Euclid.  The  majority  of 
modern  geometers,  however,  have  substituted  the  following  as 
stated  by  Newcomb,  viz. : 

^'A  straight  line  is  one  which  has  the  same  direction 
throughout  its  whole  length.'^ 

Each  is  designed  to  express  the  idea  of  straightness,  and 
not  to  convey  it ;  for  it  is  assumed  that  the  idea  already  exists 
in  the  pupil's  mind  prior  to  the  beginning  of  this  study. 

A  straight  line  may  also  be  defined  as  an  undeviating  line. 

8.  A  Curved  Line,  or  simply  curve^  is  one  no  part  of  which 
is  straight. 

9.  Material  substances  have  one  or  more  faces  which  separate 
them  from  the  rest  of  space.  These  faces  are  called  surfaces, 
and  have,  obviously,  only  two  dimensions;  i.e.  length  and 
breadth. 

The  surface  considered  apart  from  the  substance  is  called  a 
geometrical  surface. 

10.  A  Plane  is  a  geometrical  surface  such  that,  if  any  two 
points  in  it  be  selected  at  random,  the  straight  line  joining 
them  will  lie  wholly  in  that  surface. 

11.  A  Curved  Surface  is  a  geometrical  surface  no  portion  of 
which  is  a  plane. 

12.  A  physical  solid  is  the  material  composing  it,  and  which 
we  perceive  through  the  medium  of  the  senses ;  while  the 
geometrical  solid  is  the  space,  simply,  which  the  physical 
solid  occupies. 

13.  A  Geometrical  Figure  is  the  term  applied  to  combinations 
of  points,  lines,  and  surfaces,  when  reference  is  had  to  their 
form  or  outline  simply. 


Introduction.  3 

14.  A  Plane  Figure  is  one  whose  points  and  lines  all  lie  in 
the  same  plane. 

15.  "A  Plane  Rectilineal  Angle  is  the  inclination  of  two 
straight  lines  to  one  another,  which  meet  together,  but  are 
not  in  the  same  straight  line." — Euclid. 

"  An  Angle  is  a  figure  formed  by  two  straight  lines  drawn 
from  the  same  point." — Chauvenet. 

"  When  two  straight  lines  meet  together,  their  mutual  inclina- 
tion, or  degree  of  opening,  is  called  an  angleJ^  —  Loomis. 

16.  The  lines  that  form  an  angle  are  called  the  sides  of  an 
angle,  and  the  point  from  which  they  are  drawn  is  called  the 
veHex  of  the  angle. 

17.  When  two  plane  angles  have  the  same  vertex  and  a 
common  side,  neither  angle  being  a  part  of  the  other,  they  are 
said  to  be  adjacent  angles. 

18.  When  two  angles  have  the  same  vertex  and  the  sides  of 
one  are  the  extensions  of  the  sides  of  the  other,  they  are  called 
vertical  angles. 

19.  An  angle  is  named  by  a  letter  or  number  placed  at  its 
vertex.  If,  however,  there  are  two  or  more  angles  with  the 
same  vertex,  other  letters  are  placed  at  the  extremities  of  their 
sides,  and  the  three  letters  are  used  to  name  the  angle,  the 
letter  at  tJie  vertex  always  coming  between  the  other  two. 

n    1>  D 


C     ^ 


ABC 
Fig.  I.  Fig.  II.  Fio.  III. 

20.  Let  us  consider  the  point  B,  in  the  straight  line  AC, 
a  pivot,  and  BD  another  starting  from  the  position  BC,  and 
turning  about  B,  keeping  always  in  the  same  plane.     It  is 


4  Plane  Geometry. 

evident  that,  as  soon  as  it  has  started,  it  forms  tivo  angles 
with  the  line  AG^  of  which  DBC,  Fig.  I,  is  the  smaller.  If  it 
continues  to  revolve,  however,  it  will  finally  reach  a  position 
as  DB,  Fig.  II,  in  which  the  angle  DBC  is  the  larger.  Hence, 
in  passing  from  the  first  position  to  the  second,  it  must  have 
reached  a  position.  Fig.  Ill,  where  the  two  angles  DBC  and 
DBA  were  equal ;  hence, 

21.  When  one  straight  line  meets  another  so  as  to  form 
equal  adjacent  angles,  each  of  the  angles  is  called  a  right 
angle,  and  the  lines  are  said  to  be  perpendicular  to  each  other. 

22.  To  say  that  an  angle  is  a  right  angle  therefore  expresses 
the  same  fact  as  to  say  that  two  lines  are  perpendicular  to  each 
other.  The  equivalence  of  these  two  statements  should  be 
clearly  noted  and  emphasized.  It  should  also  be  borne  in 
mind  that  the  word  perpendicular  expresses  a  mutual  relation 
between  two  lines,  and  therefore  a  right  angle  may  be  defined 
as  an  angle  whose  sides  are  perpendicular  to  each  other. 

23.  It  is  evident  that  the  sum  of  the  angles  formed  by  any 
one  position  of  the  line  BD  is  equal  to  the  sum  of  the  angles 
formed  by  any  other  position;  for  what  is  taken  from  one 
angle  by  the  revolution  of  the  line  BD  is  added  to  the  other ; 
hence, 

24.  When  one  straight  line  meets  another  so  as  to  form  two 
angles,  the  sum  of  these  two  angles  equals  two  right  angles. 

25.  It  is  also  evident  that  there  can  only  be  one  such  posi- 
tion of  the  line  BD ;  hence. 

From  a  point  in  a  straight  line  only  one  perpendicular  to 
that  line  can  be  drawn,  or  if  more  be  drawn  they  will  all 
coincide. 

26.  When  one  straight  line  meets  another  so  as  to  form 
unequal  adjacent  angles,  the  lines  are  said  to  be  oblique  to 
each  other. 


Introduction.  5 

It  should  be  clearly  borne  in  mind  by  the  pupil  that  the 
word  oblique  also  in  this  connection  expresses  a  mutual  relar 
tion  between  the  two  lines. 

27.  An  angle  that  is  less  than  a  right  angle  is  called  an 
acute  angle. 

28.  An  angle  that  is  greater  than  one  right  angle  and  less 
than  two  is  called  an  obtuse  angle. 

29.  Both  acute  and  obtuse  angles  are  designated  as  oblique 
angles  as  contrasted  with  right  angles. 

30.  In  some  discussions  it  is  convenient  to  use  other  kinds 
of  angles ;  i.e. : 

A  straight  angle  is  defined  as  an  angle  whose  sides  extend 
in  exa^ly  opposite  directions  from  the  vertex. 

31.  Again,  it  is  sometimes  convenient  to  regard  an  angle  as 
formed  by  a  line  revolving  about  one  end  as  a  pivot. 

A  complete  revolution  of  such  a  line  is  called  a  perigon,  or 
round  angle. 

A  reflex  angle  is  one  that  is  larger  than  a  straight  angle  and 
smaller  than  a  perigon. 

32.  When  the  sum  of  two  angles  is  equal  to  two  right 
angles,  they  are  said  to  be  supplemental;  i.e.  each  is  the 
supplement  of  the  other. 

33.  When  the  sum  of  two  angles  is  equal  to  one  right  angle, 
they  are  said  to  be  complemental ;  i.e.  each  is  the  complement 
of  the  other. 

34.  When  the  sum  of  two  angles  equals  a  perigon,  they  are 
said  to  be  explementaX;  i.e.  each  is  the  explement  of  the 
other. 

35.  It  is  evident  from  23  and  24  that  when  one  straight  line 
meets  the  other  so  as  to  form  two  angles,  these  angles  are 
supplemental. 


6  Plane  Geometry. 

36.  Two  straight  lines  are  said  to  be  parallel  when,  lying 
in  the  same  plane,  and  extended  indefinitely  both  ways,  they 
never  meet  each  other. 

37.  As  we  have  before  conceived  a  line  (20)  to  move,  so  we 
may  conceive  one  geometrical  magnitude  to  be  applied  to  an- 
other for  the  purpose  of  comparison.  If  they  coincide,  point 
for  point,  they  are  said  to  be  congruent. 

38.  It  is  evident  that  two  geometrical  magnitudes  that  are 
congruent  are  equal  in  all  respects ;  hence, 

39.  If  two  angles  can  be  so  placed  that  their  vertices  coin- 
cide in  position  and  their  sides  in  direction,  two  and  two,  the 
angles  are  equal. 

40.  If  two  angles  are  equal  and  one  be  applied  to  the  other 
so  that  their  vertices  and  one  pair  of  sides  coincide,  then  the 
other  pair  of  sides  will  also  coincide. 

41.  Thus,  if  one  right  angle  be  applied  to  another  so  that 
their  vertices  coincide  and  likewise  one  pair  of  sides,  then, 
since  by  Sec.  25  there  can  be  only  one  perpendicular  drawn  to 
a  line  from  a  given  point  in  it,  the  other  two  sides  will  also 
coincide;  hence, 

42.  All  right  angles  are  equal. 

43.  The  word  coincide,  when  used  to  express  the  relation  of 
two  lines,  does  not  mean  that  they  necessarily  have  the  same 
extremities ;  while  the  word  congruent  does  mean  that. 

44.  Geometrical  Magnitudes  are  geometrical  lines,  angles,  sur- 
faces, and  solids. 

45.  We  shall  have  occasion  to  express  the  addition  and  sub- 
traction of  geometrical  magnitudes  as  well  as  the  multiplication 
and  division  of  these  magnitudes  by  numbers. 

46.  For  example,  the  sum  of  the  two  lines  AB  and  CD  is 
obtained  by  conceiving  them  to  be  placed  so  as  to  form  one 


Axioms.  7 

continuous  straight  line  as  HK.  Similarly,  the  difference  of 
two  lines  is  obtained  by  cutting  off  from  the  larger  a  line  equal 
to  the  smaller.  Similarly,  HKN  represents  the  sum  of  the 
two  angles  A  and  B.  To  multiply  a  line  by  a  number  is  to 
add  it  to  itself  the  required  number  of  times. 


'N 


N 


To  divide  a  line  by  a  number  is  to  conceive  the  line  to  be 
divided  into  the  required  number  of  equal  parts. 
The  same  is  true  of  other  geometrical  magnitudes. 

47.  An  Axiom  is  a  truth  that  needs  no  argument;  i.e.  the 
mere  statement  of  it  makes  it  apparent,  e.g. : 

48.  GENERAL   AXIOMS. 

i.   The  whole  of  anything  is  greater  than  any  one  of  its 
parts. 

ii.   The  whole  of  anything  is  equal  to  the  sum  of  all  its 
parts. 

ill.   Magnitudes  which   are   equal   to  the   same   or   equal 
magnitudes  are  equal  to  each  other. 

iv.   Magnitudes   which   are  halves  of  the  same  or  equal 
magnitudes  are  equal  to  each  other. 

V.   Magnitudes  which  are  doubles  of  the  same  or  equal 
magnitudes  are  equal  to  each  other. 

vi.   If  the   same   or   equal   quantities  be  added  to  equal 
quantities,  the  sums  are  equal. 

vii.   If  the   same  or  equal  quantities  be  subtracted  from 
equal  quantities,  the  remaining  quantities  are  equal. 


8  Plane  Geometry. 

viii.  If  equal  quantities  be  multiplied  by  the  same  or  equal 
quantities,  the  products  are  equal. 

ix.   If  equal  quantities  be  divided  by  the  same  or  equal 
quantities,  the  quotients  are  equal. 

X.   If  equal  quantities  be  added  to  unequal  quantities, 
the  results  will  be  unequal  in  the  same  order. 

xi.   If  equal  quantities  be  subtracted  from  unequal  quan- 
tities, the  results  will  be  unequal  in  the  same  order. 

xii.  If  unequal  quantities  be  subtracted  from  equal  quan- 
tities, the  results  will  be  unequal  in  the  reverse  order. 

xiii.   If  unequal  quantities  be  multiplied  by  the  same  or 
equal  quantities,  the  products  will  be  unequal  in  the  same  order. 
xiv.   If  unequal  quantities  be  divided  by  the  same  or  equal 
quantities,  the  quotients  will  be  unequal  in  the  same  order. 

XV.  Of  two  unequal  quantities,  a  half  of  the  larger  is 
greater  than  a  half  of  the  smaller. 

xvi.  If  unequals  be  added  to  unequals,  the  lesser  to  the 
lesser,  and  the  greater  to  the  greater,  the  sums  will  be  unequal 
in  the  same  order. 

xvii.   If  two  magnitudes  are  equal,  either  may  be  substituted 
for  the  other. 

xviii.   In  an  inequality  a  larger  quantity  may  be  substituted 
for  the  larger  and  a  smaller  for  a  smaller. 


49.  PARTICULAR  AXIOMS. 

xix.   Between  two  points  only  one   straight  line  can  be 
drawn ;  or  if  others  be  drawn,  they  will  all  coincide. 

XX.   A  straight  line  is  the  shortest  line  connecting  two 
points. 

xxi.   Conversely,  the  shortest  line  between  two  points  is  a 
straight  line. 

xxii.    If  two  straight  lines  have  two  points  in  common,  they 
will  coincide  however  far  extended. 

xxiii.   Two  straight  lines  can  intersect  in  only  one  point. 


Axioms.  9 

xxiv.  Through  a  given  point  (as  F)  only  one  line  (as  AB) 
can  be  drawn  parallel  to 

another  line  (as  CD);  or  P  b 

if  others  be  drawn,  they     c — ■ '-^ 

will  all  coincide. 

XXV.  If  a  line  makes  an  angle  with  one  of  two  parallel  lines, 
it  will  intersect  the  other  if  sufficiently  extended.  (See  above 
diagram.) 

xxvi.  The  extension  or  shortening  of  the  sides  of  an  angle 
does  not  change  the  magnitude  of  the  angle. 

50.  A  Theorem  is  a  truth  which  is  made  apparent  by  a  course 
of  reasoning  or  argument.  This  argument  is  called  a  Demon- 
stration. 

51.  Every  theorem  consists  of  two  distinct  parts,  either 
expressed  or  implied ;  viz.  the  hypothesis  and  conclusion.  The 
conclusion  is  the  part  to  be  proven,  and  the  demonstration  is 
undertaken  only  upon  the  ready  granting  of  the  conditions 
expressed  in  hypothesis ;  e.g.  : 

Hyp.     If  two  parallel  lines  be  crossed  by  a  transversal. 
Con.     the  alternate  interior  angles  are  equal. 

52.  In  demonstrating  the  theorems  in  this  book  the  pupil 
should  first  analyze  the  theorem  and  write  it  after  the  above 
model.  For  instance,  let  us  analyze  the  following  theorem; 
viz. : 

A  perpendicular  is  the  shortest  line  from  a  point  to  a  straight 
line. 

Analyzed  and  written  according  to  our  model,  this  theorem 
would  read  as  follows : 

Hyp.  If  from  a  given  point  to  a  given  straight  line  a  per- 
pendicular and  other  lines  be  drawn. 

Con.    the  perpendicular  is  the  shortest  one  of  those  lines. 

53.  The  converse  of  a  theorem  is  another  theorem  in  which 
the  hypothesis  of  the  first  becomes  the  conclusion  of  the  second, 
and  the  conclusion  of  the  first  becomes  the  hypothesis  of  the 


lO 


Plane  Geometry. 


second.    For  example,  the  converse  of  the  theorem  mentioned 
in  Sec.  51  would  xead  as  follows,  viz. : 

Hyp.     If  two  straight  lines  in  the  same  plane  be  crossed  by 
a  transversal  so  as  to  make  the  alternate  interior 
angles  equal, 
Con.     these  two  straight  lines  will  be  parallel. 

54.  A  Corollary  is  a  theorem  which  is  easily  proved  from  a 
preceding  one  with  which  it  is  closely  associated.  The  author 
deems  this  classification  of  theorems  unnecessary,  and  so  the 
term  will  be  seldom  used  in  this  book. 

55.  A  Scholium  is  a  remark  made  upon  one  or  more  preced- 
ing propositions  pointing  out  their 
application  or  limitation. 

56.  A  Problem,  in  geometry,  is  the 
required  construction  of  a  geometrical 
figure  from  stated  conditions  or  data; 
e.g.  : 

It  is  required  to  construct  the  tri- 
•^    angle  which  has  for  two  of  its  sides 
AB  and  CZ),  and  the  angle  H  included  between  these  two 
sides. 

57.  A  Postulate  is  a  self-evident  problem,  or  a  construction 
to  the  possibility  of  which  assent  is  requested  without  argu- 
ment or  evidence. 

Both  theorems  and  problems  are  commonly  designated  as 
propositions. 

58.  POSTULATES. 

Before  we  can  accomplish  the  demonstration  of  a  theorem 
in  geometry,  the  following  postulates  must  be  granted,  viz. : 

i.   A  straight  line  can  be  drawn  from  one  point  to  any  other 
point. 

ii.  A  straight  line  can  be  extended  to  any  length,  or  termi- 
nated at  any  point. 


Postulates.  1 1 

iii.  A  circle  may  be  described  about  any  point  as  a  center 
and  with  any  radius. 

iv.  Geometrical  magnitudes  of  the  same  kind  may  be  added, 
subtracted,  multiplied,  and  divided. 

V.  A  geometrical  figure  may  be  conceived  to  be  moved  at 
pleasure  without  changing  its  size  or  shape. 

59.  The  term  postulate  may  also  be  used,  and  in  this  book^ 
is  so  used,  to  designate  the  first  step  toward  the  demonstration 
whereby  it  is  requested  that  certain  conditions  be  admitted  as 
true,  for  the  basis  of  the  argument,  and  which  begins  with 
"Let,"  etc.,  or  "Let  it  be  granted  that,"  etc.  It  requests 
assent  to  the  general  conditions  implied  in  the  hypothesis, 
with  special  reference  to  a  particular  diagram,  e.g.  in  begin- 
ning the  demonstration  of  the  theorem  given  in  Sec.  51  we 
should  say,  "  Let  AB  and  HK  be 
two  parallel  straight  lines  crossed 
by  the  transversal  CX>." 

This  is  the  postulate;  i.e.  it  mat- 
ters not  whether  the  material  lines 
AB  and  HK  are  actually  straight 
or  actually  parallel ;  they  stand  for 
straight  lines  and  parallel  lines,  and  the  argument  is  just  as 
conclusive  when  based  upon  their  supposed  parallelism  as  it 
would  be  if  we  had  positive  knowledge  that  those  two  identi- 
cal lines  were  parallel. 

60.  The  demonstration  of  the  following  theorems  should  be 
written  by  the  pupils,  with  an  occasional  oral  exercise  partici- 
pated in  by  the  entire  class,  each  member  in  turn  contributing 
a  single  link  in  the  chain  of  argument. 

61.  In  order  to  save  time  for  the  pupil  in  writing  and  the 
instructor  in  correcting,  the  author,  having  used  them,  recom- 
mends the  use  of  the  following  list  of  symbols  and  abbrevia- 
tions, as  well  as  such  others  as  the  instructor  and  pupils  may 
agree  upon : 


12 


Plane  Geometry. 


SYMBOLS. 


+  .  .  .  .  plus. 

minus. 

multiplied  by. 

equals,  or  is  equal  to. 

therefore,  or  hence. 

II     parallel. 

lis parallels. 

Z angle. 

A angles. 

rt.  Z  or  Zr.  .  right  angle, 
rt.  A  or  Zr.'s    right  angles. 
A triangle. 


>  ....  is  greater  than. 
<  ....  is  less  than. 
=  or  X   equivalent  to. 
O  .  .  .  .  circle. 
CD  ...  .  circles. 

A triangles. 

rt.  A right  triangle. 

rt.  A right  triangles. 

J- perpendicular. 

Js perpendiculars. 

O parallelogram. 

HJ parallelograms. 


ABBREVIATIONS. 


Adj. 

Alt.. 

Ax.  . 

Comp. 

Con. 

Cons. 

Def. 

Dem. 

Dist. 

Ext. 

Hyp. 

Iden. 


adjacent. 

alternate. 

axiom. 

complementary. 

conclusion. 

construction. 

definition. 

demonstration. 

distance. 

exterior. 

hypothesis. 

identical. 

Q.E.D.  . 

Q.E.r.  . 


Quod 
Quod 


Int interior. 

Line straight  line. 

Opp opposite. 

Post postulate. 

Prob problem. 

Pt point. 

Quad quadrilateral. 

St straight. 

Sug suggestion. 

Sup supplementary. 

Trans transversal. 

Vert vertical. 

erat  demonstrandum, 
erat  faciendum. 


The  last  two  expressions  are  in  Latin,  and  mean,  respec- 
tively, "which  was  to  be  demonstrated"  or  "proven,"  and 
"which  was  to  be  performed"  or  "done."  The  former  is 
placed  at  the  close  of  the  demonstration  of  every  theorem  to 
indicate  that  the  required  proof  has  been  completed,  while 


Symbols  and  Abbreviations.  13 

the  latter  is  placed  at  the  close  of  the  work  of  every  problem 
to  indicate  that  the  required  construction  has  been  accom- 
plished. 

62.  The  pupil  must  remember  that  evei^  statement  in  geo- 
metrical demonstration  must  be  "  backed  up  "  or  substantiated 
by  giving  as  authority  definitions,  axioms,  and  previously 
established  truths,  and  every  unsupported  statement ,  whether 
from  instructor  or  fellow-pupil,  should  be  promptly  challenged. 

63.  The  author  here  wishes  to  emphasize  this  caution  to 
both  instructor  and  pupil,  based  on  several  years'  experience 
in  the  class-room,  viz.:  always  employ  the  most  unfavorable 
diagram.  Care  in  this  particular  will  save  many  an  inad- 
vertent error  and  prevent  the  assumption  of  conditions  unwar- 
ranted by  the  hypothesis. 


14  Plane  Geometry. 

THEOREMS. 

64.  If  two  angles  are  equal,  their  complemental  angles  are  also 
equal. 

65.  If  an  angle  is  complemental  to  each  of  two  other  angles, 
these  other  two  angles  are  equal. 

66.  If  two  angles  are  equal,  their  supplemental  angles  are  also 
equal. 

67.  If  an  angle  is  supplemental  to  each  of  two  other  angles, 
these  other  two  angles  are  equal. 

68.  If  two  supplemental  angles  are  equal,  each  is  a  right 
angle. 

69.  If  one  straight  Hue  intersects  another,  the  vertical  angles  are 
equal. 

Sug.     Consult  67. 

70.  If  a  line  bisect  an  angle,  and  a  perpendicular  be  drawn 
to  this  bisector  through  the  vertex,  the  angles  which  this 
perpendicular  makes  with  the  sides  of  the  angle  are  equal. 

71.  If  two  straight  lines  intersect  each  other,  the  sum  of  the 
four  angles  equals  four  right  angles. 

72.  The  sum  of  all  the  consecutive  angles  formed  at  one  point 
in  a  straight  Hne  and  on  one  side  of  it  equals  two  right  angles. 

Sug.     Draw  a  perpendicular,  or  consult  24. 

73.  If  any  number  of  lines  be  drawn  from  the  same  point, 
the  sum  of  all  the  consecutive  angles  equals  four  right  angles. 

Sug.     Extend  one  of  the  lines  from  the  point. 

74.  If  one  of  the  angles  formed  by  the  intersection  of  two 
straight  lines  is  a  right  angle,  the  other  three  are  also  right 
angles. 

Sug.     Consult  69  and  24. 

75.  If  two  adjacent  supplemental  angles  are  bisected,  the  bisectors 
are  perpendicular  to  each  other. 


Angles.  1 5 

Sug.  Prove  that  the  angle  which  they  form  with  each  other 
is  a  right  angle. 

76.  If  the  bisectors  of  two  adjacent  angles  are  perpendicular  to  each 
other,  the  two  angles  are  supplemental. 

77.  If  one  of  two  adjacent  supplemental  angles  be  bisected, 
and  a  perpendicular  be  drawn  to  this  bisector  from  the  common 
vertex,  this  perpendicular  will  bisect  the  other  angle. 

78.  If  two  adjacent  angles  are  supplemental,  their  exterior  sides 
form  one  straight  Hue. 

Post.  Let  ABH  and  BBC  be  2  adj.  4  and  also  let  Z  ABH 
-^ZHBC  =  2  It.  A. 


To  Prove.    AB  is  the  extension  of  BC. 

Cons.    Draw  BD  to  represent  the  extension  of  BC. 

Dem. 


Z  ABH+  A  HBC=  2  rt.  A. 

(?) 
Z  DBH-\-  Z  HBC  =  2  rt.  A. 

(?) 
.'.ZABH^-ZDBH; 

(?) 


.-.  AB  must  coincide  with  BD. 

(?) 

.*.  AB  is  the  extension  of  BC 

(?) 

i.e.  AB  and  BC  form  one  line. 

Q.E.D. 


79.  If  two  vertical  angles  be  bisected,  the  bisectors  will  form 
one  line. 

Sug.     Consult  72  and  78,  or  draw  a  bisector  of  one  of  the 
other  angles,  and  consult  76  and  25. 

80.  If  a  line,  bisecting  one  of  two  vertical  angles,  be  extended  from 
the  vertex,  this  extension  will  bisect  the  other  angle. 

Sug.     Consult  69  and  Ax.  iii.,  or  draw  a  perpendicular  from 
the  vertex,  and  consult  77. 


1 6  Plane  Geometry. 


TRIANGLES. 

81.  A  Triangle  is  a  plane  figure  bounded  by  three  straight 
lines,  and  consequently  has  three  angles.  It  is  sometimes 
called  a  trigon. 

82.  Triangles  are  named, 

i.   With  reference  to  the  character  of  their  angles. 
ii.   With  reference  to  the  relations  between  their  sides. 

i.   If  a  triangle  has  all  its  angles  acutCj  it  is  called  an 
acute  triangle. 

If  it  has  one  obtuse  angle,  it  is  called  an  obtuse  triangle. 

If  it  has  one  right  angle,  it  is  called  a  inght  triangle. 
ii.   If  all  its  sides  are  equal,  it  is  called  an  equilateral  tri- 
angle. 

If  two  of  its  sides  are  equal,  it  is  called  an  isosceles  tri- 
angle. 

If  no  two  of  its  sides  are  equal,  it  is  called  a  scalene 
triangle. 

83.  The  term  isosceles  is  used  to  designate  the  fact  that  two 
sides  are  equal  without  any  reference  to  the  third  side,  even 
though  it  may  then  be  known,  or  afterward  ascertained,  that 
the  sides  are  all  equal. 

84.  If  the  angles  of  a  triangle  are  all  equal,  the  triangle  is 
said  to  be  equiangular. 

If  the  three  angles  of  one  triangle  are  equal  respectively  to 
the  three  angles  of  another,  the  two  triangles  are  said  to  be 
mutually  equiangular. 

85.  If  the  sides  of  one  triangle  are  equal  respectively  to  the 
three  sides  of  another,  the  two  triangles  are  said  to  be  mutually 
equilateral. 

86.  The  corresponding  sides  and  angles  of  two  triangles  are 
those  that  are  similarly  situated,  and  are  called  homologous. 
If  the  triangles  are  equal,  the  homologous  sides  are  those  that 


Triangles.  *  17 

are  opposite  the  equal  angles ;  and  conversely,  the  homologous 
angles  are  those  that  are  opposite  the  equal  sides. 

87.  The  sum  of  the  three  sides  of  a  triangle  is  called  the 
perimeter. 

In  a  right  triangle,  the  side  opposite  the  right  angle  is  called 
the  hypothenuse,  and  the  other  two  sides  the  legs. 

The  side  on  which  a  triangle  is  supposed  to  rest  is  called  the 
base.  In  general,  any  side  of  a  triangle  may  be  considered  the 
base ;  but  in  an  isosceles  triangle  the  third  side,  and  in  a  right 
triangle  one  of  the  legs,  is  generally  the  base. 

The  angle  opposite  the  base  is  called  its  vertical  angle ;  and 
its  vertex,  the  vertex  of  the  triangle. 

The  perpendicular  distance  from  vertex  to  base,  or  base  ex- 
tended, is  called  the  altitude  of  a  triangle. 

The  line  drawn  from  the  vertex  of  a  triangle  to  the  middle 
point  of  the  base  is  called  a  median. 

88.  If  two  or  more  lines  meet- or  intersect  one  another  at  a 
common  point,  they  are  said  to  be  concuiirent,  and  if  two  or  more 
points  lie  in  the  same  straight  line  they  are  said  to  be  collinear. 

THEOREMS. 

89.  Any  side  of  a  triangle  is  less  than  the  sum  of  the  other  two. 

90.  Any  side  of  a  triangle  is  greater  than  the  difference  of  the 
other  two. 

Sug.     Consult  89  and  Ax.  vii. 

91.  If  lines  be  drawn  from  a  point  within  a  triangle  to  the  ends  of 
either  side,  the  sum  of  these  two  lines  is  less  than  the  sum  of  the  other 
two  sides  of  the  triangle. 

Sug.  Produce  one  of  the  lines  until  it  meets  a  side  of  the 
triangle,  then  consult  89  and  Ax.  xvi. 

92.  If  two  sides  and  included  angle  of  one  triangle  are  equal  re- 
spectively to  two  sides  and  included  angle  of  another,  the  two  triangles 
are  equal. 


1 8  '  Plane  Geometry. 

Sug.  Superpose  one  triangle  upon  the  other  so  that  they 
will  have  a  common  vertex  and  a  common  side,  then  prove  the 
congruence  of  the  other  parts.     See  37  and  38. 

93.  If  two  sides  of  a  triangle  are  equal,  the  angles  opposite  those 
sides  are  also  equal. 

Sug.  Draw  a  line  bisecting  the  vertical  angle,  then  compare 
the  triangles  by  92. 

94.  If  two  angles  and  included  side  of  one  triangle  are  equal  re- 
spectively to  two  angles  and  included  side  of  another,  the  two  triangles 
are  equal. 

Sug.     Use  the  same  method  as  in  92. 

95.  If  two  triangles  are  mutually  equilateral,  they  are  equal. 
Sug.    Apply  one  to  the  other  as  in  Fig.  I,  and  consult  93 

and  92,  or,  as  in  Fig.  II,  and  prove  that  point  G  must  fall  on 
point  H. 


y^  Fig.  I.  Fig.  II. 

96.  If  two  angles  of  a  triangle  are  equal,  the  sides  opposite  those 
angles  are  also  equal. 

Sug.  Draw  a  line  from  the  vertex  to  the  center  of  the  base 
and  compare  the  triangles. 

97.  If  two  angles  of  a  triangle  are  unequal,  the  side  opposite  the 
larger  angle  is  longer  than  the  side  opposite  the  lesser. 

Sug.  Take  a  part  of  the  larger  angle  equal  to  the  smaller 
angle  and  have  them  both  in  the  same  triangle,  then  consult 
96  and  89. 


Triangles. 


19 


98.   From  a  point  to  a  straight  line  only  one  perpendicular  can  be 
drawn  to  that  line,  or  if  others  are  drawn  they  will  all  coincide. 


Post.  Let  AC  be  ±  to  BF 
and  AD  any  other  line  from 
A  to  BF. 

To  Prove.  AD  is  oblique  to 
BF. 

Cons.  Prolong  AC  from  C 
making  CH=AC,  join  HD 
and  extend  AD  from  D. 


Dem. 

AACD  =  ACDR. 

(?) 
.'.  Z  ADC  =  Z  CDir. 

(?) 

ZCDH<ZCDW, 

(?) 
etc. 


99.  A  perpendicular  is  the  shortest  line  that  can  be  drawn  from  a 
point  to  a  straight  line. 


Cms.    Draw  DN  ±  to  BF. 
A        N 


D 


Dem.     AD    is    oblique    to 
BF.     (?) 
.-.  ADO  is  an  acute  Z.     (?) 


ZACD  =  ZCDN. 

(?) 
ZCDN>ZADC. 

(?) 
ZACD>ZADC. 

.-.  AC<AD. 

(?)  Q.E.D. 


20 


Plane  Geometry. 


100.  If  a  perpendicular  bisect  a  line,  tlie  lines  drawn  from  any 
point  in  the  perpendicular  to  the  ends  of  the  line  are  equal ;  i.e.  the 
point  is  equidistant  from  the  ends  of  the  line, 

Sug.     Consult  92. 

101.  If  each  of  two  points  be  equidistant  from  the  ends  of  a  given 
line,  the  line  passing  through  these  points  will  be  the  perpendicular 
bisector  of  the  given  line. 

Sug.     Consult  95  and  92. 

102.  If  a  perpendicular  bisect  a  line,  and  lines  be  drawn  from 
any  point  outside  this  perpendicular  to  the  ends  of  the  line,  the  line 
that  crosses  the  perpendicular  is  the  longer. 

Sug.  Draw  a  line  from  the  point  of  intersection  so  as  to 
utilize  96  and  89. 

103.  Every  point  which  is  equidistant  from  the  ends  of  a  given 
line  is  in  the  perpendicular  bisector  of  that  line. 

104.  Only  two  equal  lines  can  be  drawn  from  a  point  to  a  straight 
line. 

105.  The  shortest  line  from  a  point  to  a  line  is  perpendicular 
to  the  line. 

Post.  Let  CD  be  the  short- 
est line  from  C  to  AB. 

To  Prove.    CD  is  J_  to  AB. 

Dem.  CD  is  either  J_  to  AB 
or  it  is  not.  Let  us  suppose  it 
A 


B 

is  not.  Then  draw  CF  J.  to  AB 
and  produce  it  from  F,  making 
FH=  FC,  and  join  HD. 


HD-\-DC>HF+FC. 

(?) 
HD=DC. 

(?) 
EF=FG. 

(?) 
.-.  DC>FC. 

(?) 
But         DC<FC. 

(?) 
.*.  we  have  arrived  at  a  re- 
sult   which    contradicts     the 
Hyp.  etc.  Q.E.D. 


Transversals.  21 

106.  If  a  perpendicular  divide  a  line  unequally  and  lines  be  drawn 
from  any  point  in  the  perpendicular  to  the  ends  of  the  line,  the  line 
drawn  to  the  end  the  farther  from  the  perpendicular  is  the  longer. 

Sug.  Draw  the  ±  bisector  of  the  line,  and  draw  another 
line  so  as  to  utilize  96  and  97. 

107.  If  two  sides  of  a  triangle  are  unequal,  the  angle  opposite 
the  longer  side  is  greater  than  the  angle  opposite  the  shorter. 

Sug.  Draw  the  J_  bisector  of  the  third  side,  and  connect 
the  point  of  intersection  and  end  of  the  third  side.  Then  con- 
sult 92  and  93. 

108.  If  two  lines  are  perpendicular  to  the  same  line,  they  are 
parallel. 

/Stcg.  Two  possible  relations.  One  of  them  contradicts  a 
previous  theorem. 

109.  If  a  Hue  is  perpendicular  to  one  of  two  parallel  lines,  it  is  per- 
pendicular to  the  other  also. 

Sug.  From  the  point  of  intersection  with  the  latter  draw 
a  ±  to  the  perpendicular,  then  consult  108  and  Ax.  xxiv. 

110.  If  two  Hues  are  parallel  to  the  same  line,  they  are  parallel  to 
each  other. 

Sug.    Draw  a  ±  to  one  of  the  lines,  and  consult  109. 

TRANSVERSALS. 

111.  When  two  straight  lines  in  the  same  plane  are  crossed 
by  another,  the  latter  is  called  a  transversal.  In  the  figure, 
which  line  is  the  transversal  ? 

How    many    angles     are  ^ 

formed?      Certain   ones    of     X-^ 

the  above  angles  are  termed 
interior  angles.  Name  them. 
Why  are  they  called  interior 
angles  ? 

What    name    would    you 


22 


Plane  Geometry. 


give  to  the  others  to  distinguish  them  from  those  already- 
named  ?  Why  ?  How  many  pairs  of  vertical  angles  ?  Name 
them. 

How  many  pairs  of  adjacent  angles  ?    Name  them. 

Two  angles  situated  on  opposite  sides  of  the  transversal, 
either  both  interior  or  both  exterior,  and  not  adjacent,  are 
called  alternate  angles. 

How  many  pairs  of  alternate  interior  angles  ?    Name  them. 

How  many  pairs  of  alternate  exterior  angles  ?    Name  them. 

Two  non-adjacent  angles,  of  which  one  is  interior  and  the 
other  exterior,  and  both  on  the  same  side  of  transversal,  are 
called  technically  exterior  interior  angles. 

How  many  pairs  of  exterior  interior  angles  ?    Name  them. 

Angles  on  the  same  side  of  the  transversal  are  called  lateral 
angles. 

How  many  pairs  of  lateral  interior  angles  ?     Name  them. 

How  many  pairs  of  lateral  exterior  angles  ?   Name  them. 

112.  If  two  lines  in  the  same  plane  are  crossed  by  a  transversal 
making  the  alternate  interior  angles  equal,  the  two  lines  are  parallel, 

Post.  Let  AB  and  CD  be  two  lines  crossed  by  trans.  FH, 
making  Z  CNK=  Z.  NKB. 


R 


To  Prove.    AB  is  II  to  CD. 

Cons.     Bisect  NK  and  through  its  center  0  draw  RP  JL 
to  CD. 
Dem.    Compare  the  A,  and  consult  108. 


Transversals. 


23 


A 

■b:/ 

Ti 

W 

^ 

f 

C         /S 

4 

Post.     

113.  If  two  parallel  lines  are  crossed  by  a  transversal,  tlie  alternate 
interior  angles  are  equal. 

Dem. 
AWKN=Z.KND. 

(?) 
.-.  WK\?>  II  to  CD, 

(?) 
AB  is  II  to  CD. 

(?) 
.-.  AB   coincides  with  WK 
10  I^rove.  .  Qj.  contradicts  Ax.  xxiv,  etc. 

Cons.     Draw    WKj   making  q  e  d 

ZWKN=ZKND. 

114.  If  two  parallel  lines  are  crossed  by  a  transversal,  the  exterior 
interior  angles  are  equal. 

115.  Converse  of  114. 

116.  If  two  parallel  lines  are  crossed  by  a  transversal,  the 
alternate  exterior  angles  are  equal. 

117.  Converse  of  116. 

118.  If  two  parallel  hues  are  crossed  by  a  transversal,  the  lateral 
interior  angles  are  supplemental. 

119.  Converse  of  118. 

120.  If  two  parallel  lines  are  crossed  by  a  transversal,  the 
lateral  exterior  angles  are  supplemental. 

121.  Converse  of  120. 

122.  If  two  lines  in  the  same  plane  be  crossed  by  a  transversal 
making  the  alternate  interior  angles  unequal,  the  lines  will  meet,  if 
sufficiently  prolonged,  on  that  side  of  the  transversal  on  which  is  the 
smaller  interior  angle. 

Sug.  Draw  a  line  through  the  vertex  of  the  larger  angle 
making  an  angle  equal  to  the  smaller  so  as  to  utilize  112  and 
Ax.  XXV. 


24  Plane  Geometry. 

123.  If  two  lines  in  the  same  plane  be  crossed  by  a  transversal 
making  the  sum  of  the  lateral  interior  angles  less  than  two  right 
angles,  the  lines  will  meet,  if  sufficiently  extended,  on  that  side  of  the 
transversal  on  which  the  sum  of  the  angles  is  less  than  two  right 
angles.     Sug.     Consult  122. 

124.  If  two  parallel  lines  are  crossed  by  a  transversal,  the 
lines  bisecting  either  pair  of  alternate  interior  angles  are 
parallel. 

125.  If  two  parallel  lines  are  crossed  by  a  transversal,  the 
lines  bisecting  the  lateral  interior  angles  are  perpendicular  to 
each  other. 

Sug.     Consult  124,  75,  and  109. 

126.  The  sum  of  the  three  angles  of  a  triangle  equals  two  right 
angles.     Sug.     Consult  72  and  113. 

127.  If  two  angles  of  one  triangle  are  equal  respectively  to  two 
angles  of  another,  their  third  angles  are  equal. 

128.  If  the  sum  of  two  angles  of  a  triangle  is  equal  to  the  third, 
the  latter  is  a  right  angle. 

129.  If  one  side  of  a  triangle  be  produced,  the  exterior  angle  equals 
the  sum  of  the  two  interior  angles  not  adjacent  to  it. 

130.  If  one  of  the  equal  sides  of  an  isosceles  triangle  be 
extended  at  the  vertex,  the  exterior  angle  is  double  either  of 
the  base  angles. 

131.  If  the  legs  of  a  right  triangle  are  equal  respectively  to 
the  legs  of  another,  the  triangles  are  equal. 

132.  If  the  hypothenuse  and  an  acute  angle  of  one  right 
triangle  are  equal  respectively  to  the  hypothenuse  and  acute 
angle  of  another,  the  triangles  are  equal. 

Sug.     Consult  127  and  94. 

133.  If  a  leg  and  acute  angle  of  one  triangle  are  equal 
respectively  to  a  leg  and  homologous  acute  angle  of  another, 
the  triangles  are  equal. 


Triangles. 


25 


Sug.    Case  I  when  the  equal  angles  are  adjacent  to  the  legs. 
Case  II  when  the  equal  angles  are  opposite  the  legs. 

134.  If  the  hypothemise  and  leg  of  one  right  triangle  are  equal 
respectively  to  the  hypothenuse  and  leg  of  another,  the  triangles  are 
equal. 


Cons.  Extend  CB  from  B  making  BN=  FH,  then  consult 
131, 103,  and  95. 

135.  If  two  sides  of  one  triangle  are  equal  respectively  to  two  sides 
of  another,  but  the  included  angle  of  one  greater  than  the  included 
angle  of  the  other,  then  the  third  side  of  that  triangle  having  the 
greater  included  angle  is  longer  than  the  third  side  of  the  other. 


Post.  Let^lBCandDFffbe 
2  A  with  sides  AB  =  DF  and 
BG  =  FH,  and  Z  ABC  greater 
than  Z  F,  and  let  AB  be  the 
side  not  greater  than  BC. 

To  Prove  AC>DH. 


Cons.  Draw  AK  and  BK 
equal  respectively  to  DH  and 
FH.  Draw  BN  bisecting 
Z  KBC  and  join  KN. 

Dem.  Compare  A  AB/^and 
DFH,  also  AKBN  and  BCNy 
then  consult  89. 


26  Plane  Geometry. 

136.  If  two  sides  of  one  triangle  are  equal  respectively  to  two 
sides  of  another,  bnt  their  third  sides  unequal,  then  the  angle  oppo- 
site the  longer  third  side  is  greater  than  the  angle  opposite  the 
shorter. 

Sug.     Three  possible  relations. 

137.  If  two  sides  and  obtuse  angle  opposite  one,  of  one 
triangle,  be  equal  respectively  to  two  sides  and  obtuse  angle 
opposite  one,  of  another,  the  two  triangles  are  equal. 

A 


w      B  CO J ^£r 

Cons.  Produce  BC  and  FH  from  B  and  F.  Draw  MAW 
and  DO  and  compare  A. 

138.  If  two  triangles  are  equal,  their  homologous  altitudes 
are  equal. 

139.  If  a  triangle  is  equilateral,  it  is  also  equiangular. 

140.  Converse  of  139. 

141.  If  a  line  bisect  the  vertical  angle  of  an  isosceles  triangle,  it 
will  be  perpendicular  to,  and  bisect,  the  base. 

142.  If  a  perpendicular  bisect  the  base  of  an  isosceles  triangle, 
this  perpendicular  will  pass  through  the  vertex  and  bisect  the  ver- 
tical angle. 

143.  If  a  line  be  drawn  from  the  vertex  of  an  isosceles  triangle 
perpendicular  to  the  base,  it  will  bisect  both  the  base  and  vertical 
angle. 

144.  If  a  line  be  drawn  from  the  vertex  of  an  isosceles  triangle 
to  the  center  of  the  base,  it  will  be  perpendicular  to  the  base  and 
bisect  the  vertical  angle. 


Bisectors. 


27 


145.  The  three  altitudes  of  an  equilateral  triangle  are  equal. 

146.  If  a  line  bisect  an  angle,  the  perpendiculars  from  any  point 
in  the  bisector  to  the  sides  of  the  angle  are  equal;  i.e.  the  point  is 
equidistant  from  the  sides  of  the  angle. 

147.  If  equal  perpendiculars  be  drawn  from  any  point  to  the  sides 
of  an  angle,  the  line  which  joins  this  point  to  the  vertex  will  bisect 
the  angle ;  i.e,  if  a  point  be  equidistant  from  the  sides  of  an  angle,  etc. 

148.  If  a  line  bisect  an  angle  and  perpendiculars  be  drawn 
from  any  point  outside  this  bisector  to  the  sides  of  the  angle, 
the  perpendicular  that  crosses  the  bisector  is  the  longer. 

jg-  B 


To  Prove.     0H>  OF. 
Cons.     Draw   KNl.  to   BC 
and  join  ON. 

Sug.    Compare  Oi^  with  OJV, 
and  OiV^with  OK+KN,  etc. 
Post.     Let  ABC   be   an  Z 
bisected  by  DB,  etc. 

149.  The  three  bisectors  of  the  angles  of  a  triangle  are  concurrent. 
Sug.     From  the  point  of  intersection  of  two  bisectors  draw  a 

line  to  the  third  vertex  and  also  Js  to  the  three  sides.  Prove 
the  former  a  bisector  by  a  comparison  of  triangles,  or  consult 
146  and  147. 

150.  The  perpendicular  bisectors  of  the  sides  of  a  triangle  are 
concurrent. 

Sug.     From  the  point  of  intersection  of  two  of  the  Js  draw  a 
line  to  the  center  of  the  third  side  and  prove  this  line  a  J_. 

151.  The  perpendiculars  from  the  three  vertices  of  a  triangle  to 
the  opposite  sides  are  concurrent. 

Sug.    Through  each  vertex  draw  a  line  I|  to  the  opposite 
side  and  consult  150. 


28  Plane  Geometry. 

ADVANCE  THEOREMS. 

The  following  theorems  are  to  he  demonstrated  entirely  indepen- 
dent of  one  another. 

They  vary  as  to  difficulty,  and  are  given  for  the  purpose  of 
affording  additional  practice  for  the  pupil  if  the  instructor 
should  deem  it  necessary,  as  well  as  offering  further  scope  for 
the  enthusiasm  of  the  pupil. 

152.  If  one  angle  of  an  isosceles  triangle  is  two  thirds  of  a 
right  angle,  the  triangle  is  equilateral. 

153.  If  the  base  angles  of  an  isosceles  triangle  be  bisected, 
the  bisectors  will  form,  with  the  base,  an  isosceles  triangle. 

154.  If  one  of  the  equal  sides  of  an  isosceles  triangle  be 
extended  at  the  vertex  and  the  exterior  angle  thus  formed 
be  bisected,  this  bisector  will  be  parallel  to  the  base. 

155.  If  one  of  the  equal  sides  of  an  isosceles  triangle  be 
extended  at  the  vertex  and  a  line  be  drawn  through  the  vertex 
parallel  to  the  base,  this  line  will  bisect  the  exterior  angle. 

156.  If  one  of  the  equal  sides  of  an  isosceles  triangle  be 
extended  at  the  vertex,  making  the  extension  equal  to  the 
side,  and  its  extremity  joined  to  the  end  of  the  base,  this  line 
will  be  perpendicular  to  the  base. 

157.  If  each  of  the  base  angles  of  an  isosceles  triangle  be 
double  the  vertical  angle,  a  line  bisecting  either  of  the  former 
will  divide  the  triangle  into  two  isosceles  triangles. 

158.  If  two  triangles  have  two  unequal  sides  and  an  angle 
opposite  the  greater  of  one,  equal  respectively  to  two  unequal 
sides  and  an  angle  opposite  the  greater  of  the  other,  the  two 
triangles  are  equal. 

Sug.  Erom  the  vertices  of  the  angles  formed  by  the  given 
sides  draw  Js  to  the  third  sides  or  their  extensions,  and  com- 
pare the  right  angles. 


Theorems. 


29 


159.  If  two  angles  of  an  equilateral  triangle  be  bisected  and 
lines  be  drawn  through  the  point  of  intersection  parallel  to  the 
sides,  the  sides  will  be  trisected. 

160.  The  two  perpendiculars  from  the  extremities  of  the 
base  of  an  isosceles  triangle  to  the  equal  sides  are  equal. 

161.  Converse  of  160. 

162.  The  medians  drawn  to  the  equal  sides  of  an  isosceles 
triangle  are  equal. 

163.  Converse  of  162. 

164.  The  bisectors  of  the  base  angles  of  an  isosceles  triangle, 
terminating  in  the  equal  sides,  are  equal. 

165.  Converse  of  164. 


Dem.  ADBK=AFAB. 

AB=DK. 
Z  AHB=Z  ADH+  Z  DAH. 
AAHB=./.ADK. 
Z.  AHB=  Z  HFB+Z  IIBF. 
Z  AHB=Z  HBK+Z  HBA. 
ZAHB=ZABK 
ZADK=^ZABK. 
ZNDK=ZABP. 
ANDK=AABP. 

]SrD=BPsind  NK=AP. 
ANAK=AAKP. 

NA=KP. 

AD=KB. 
But  FB=KB. 

AD=FB. 
AADB=AAFB. 
Z  ABD=ZFAB. 
ZDAB=ZABF. 

AC=BC.  Q.E.D. 


\       / 

Post.  Let  AF=  DB  and  bi- 
sect A  DAB  and  ABF. 

To  Prove.    AC  =  BC. 

Cons.   Make 

Z  BDK=  Z  FAB, 
and         ZDBK^ZAFB. 

Join  AK.  Extend  BK  and 
draw  API.  to  KP  and  KNA. 
to  AC. 


30  Plane  Geometry. 

166.  The  two  perpendiculars  from  the  middle  point  of  the 
base  of  an  isosceles  triangle  to  the  equal  sides  are  equal. 

167.  Converse  of  166. 

168.  If  perpendiculars  be  drawn  to  the  equal  sides  of  an 
isosceles  triangle  through  the  extremities  of  the  base  and  meet- 
ing below  the  base,  the  triangle  thus  formed  is  isosceles. 

169.  If  a  median  bisects  an  angle  of  a  triangle,  this  triangle 
is  isosceles. 

170.  If  the  vertical  angle  of  an  isosceles  triangle  is  a  right 
angle,  the  altitude  is  equal  to  one  half  the  base. 

171.  If  a  line  be  drawn  parallel  to  the  base  of  an  isosceles 
triangle  and  meeting  the  other  sides,  the  triangle  thus  formed 
is  isosceles. 

172.  If  a  line  bisecting  an  exterior  angle  of  a  triangle  is 
parallel  to  a  side,  this  triangle  is  isosceles. 

173.  If  two  isosceles  triangles  have  the  base  and  vertical 
angle  of  one  equal  to  the  base  and  vertical  angle  of  the  other, 
the  two  triangles  are  equal. 

174.  If,  from  any  point  in  the  bisector  of  an  angle,  a  line  be 
drawn  parallel  to  one  side  and  terminating  in  the  other,  the 
triangle  thus  formed  is  isosceles. 

175.  If  a  perpendicular  be  drawn  to  the  base  of  an  isosceles 
triangle  intersecting  one  of  the  equal  sides  and  produced  to  meet 
the  extension  of  the  other,  the  triangle  thus  formed  is  isosceles. 

176.  If  the  base  angles  of  an  isosceles  triangle  be  bisected, 
the  line  joining  the  points  of  intersection  of  these  bisectors 
with  the  sides  of  the  triangle  is  parallel  to  the  base. 

177.  If  medians  be  drawn  to  the  equal  sides  of  an  isosceles 
triangle,  the  line  joining  the  points  of  meeting  is  parallel  to 
the  base. 


Theorems.  3 1 

178.  If  perpendiculars  be  drawn  to  the  equal  sides  of  an 
isosceles  triangle  from  the  ends  of  the  base,  the  line  joining 
the  points  of  meeting  is  parallel  to  the  base, 

179.  If  one  side  of  a  triangle  be  produced  in  both  directions, 
the  sum  of  the  exterior  angles  exceeds  two  right  angles. 

180.  If,  from  a  point  within  a  triangle,  lines  are  drawn  to 
the  ends  of  one  side,  the  angle  formed  by  these  lines  is  greater 
than  the  angle  formed  by  the  other  two  sides  of  the  triangle. 

181.  If  the  two  base  angles  of  a  triangle  are  bisected  and 
through  the  point  of  intersection  a  line  be  drawn  parallel  to 
the  base  and  terminating  in  the  sides,  this  line  is  equal  to  the 
two  segments  of  the  sides  between  the  two  parallels. 

182.  The  bisectors  of  the  vertical  angle  of  a  triangle  and  of 
the  angles  formed  by  extending  the  sides  below  the  base  are 
concurrent,  and  the  point  of  meeting  is  equidistant  from  the 
base  and  sides  extended. 

183.  If  a  line  be  drawn  from  one  extremity  of  the  base  of  an 
isosceles  triangle  perpendicular  to  the  opposite  side,  the  angle 
formed  by  this  line  and  the  base  is  one  half  the  vertical  angle. 

184.  If  an  angle  be  bisected  and  a  perpendicular  be  drawn 
to  this  bisector  terminating  in  the  sides  of  the  angle,  the 
triangle  thus  formed  is  isosceles. 


32  Plane  Geometry. 


ANGLE  MEASUREMENT. 

185.  The  most  commoii  unit  for  measuring  the  magnitude 
of  angles  is  the  ninetieth  part  of  a  right  angle,  called  a  degree. 
The  degree  is  subdivided  into  sixty  equal  parts  called  minutes, 
and  the  latter  also  into  sixty  equal  parts  called  seconds.  By 
this  means  small  fractions  of  a  degree  may  be  expressed  in 
minutes  and  seconds.  These  units  are  indicated  by  the  follow- 
ing symbols,  viz.: 

°  for  degrees,  '  for  minutes,  and  "  for  seconds ;  e.g.  75°  20'  34". 

EXERCISES. 

186.  1.  How  many  degrees  in  a  right  angle  ? 

2.  What  is  the  complement  of  a  right  angle  ? 

3.  What  is  the  supplement  of  a  right  angle  ? 

4.  How  many  degrees  in  all  the  consecutive  angles  formed 
at  one  point  on  one  side  of  a  straight  line  ? 

5.  How  many  degrees  in  all  the  consecutive  angles  formed 
by  any  number  of  lines  meeting  at  one  point  ? 

6.  What  is  the  complement  of  45°;  30°;  1°;   80°  59'  59"; 
27°  37'  47"  ? 

7.  What  is  the  supplement  of  each  of  the  above-named 
angles  ? 

8.  How  many  degrees  in  an  angle  that  is  double  its  comple- 
ment? 

9.  How  many  degrees  in  an  angle  that  is  four  times  its 
supplement  ? 

10.  What  angle,  in  degrees,  do  the  hour  and  minute  hands 
of  a  clock  form  at  2  o'clock  ?     At  3  o'clock  ?     At  5  o'clock  ? 

11.  If  one  of  the  angles  formed  by  two  straight  lines  cross- 
ing each  other  be  120°,  what  are  the  values  of  the  other  angles  ? 

12.  If  two  complemental  adjacent  angles  be  bisected,  how 
many  degrees  in  the  angle  formed  by  the  bisectors  ? 


Angle  Measurement.  33 

13.  What  is  the  complement  of  37°  44'  51"  ? 

14.  What  is  the  supplement  of  104°  33'  21"  ? 

15.  One  half  a  right  angle  is  what  part  of  three  right 
angles?     Of  two  right  angles? 

16.  One  fifth  of  two  right  angles  is  what  part  of  one  right 
angle  ?     Of  four  right  angles  ? 

17.  How  many  degrees  in  an  angle  that  is  one  fifth  of  its 
complement  ? 

18.  How  many  degrees  in  an  angle  that  is  one  third  of  its 
supplement  ? 

19.  Two  lines  are  crossed  by  a  transversal,  making  one  pair 
of  lateral  exterior  angles  54°  48'  32"  and  125°  11' 28"  respec- 
tively; how  are  the  two  lines  related  to  each  other  ? 

20.  If  one  of  two  complemental  angles  is  acute,  what  kind 
of  an  angle  must  the  other  be  ? 

21.  If  one  of  two  supplemental  angles  is  acute,  what  kind 
must  the  other  one  be  ? 

22.  If  two  angles  of  a  triangle  be  known,  how  can  the  third 
be  found  ? 

23.  Two  angles  of  a  triangle  are  34°  28'  42"  and  29°  44'  56" 
respectively ;  find  the  value  of  the  third  angle. 

24.  If  one  angle  of  a  triangle  be  a  right  angle,  what  rela- 
tion exists  between  the  other  two  ? 

25.  How  many  right  angles  can  there  be  in  a  triangle? 
How  many  obtuse  angles  ? 

26.  If  the  three  angles  of  a  triangle  are  all  equal,  what  is 
the  value  of  each  angle  in  terms  of  a  right  angle  ?    In  degrees  ? 

27.  If  one  acute  angle  of  a  right  triangle  is  27°  38'  50",  what 
is  the  value  of  the  other  acute  angle  ? 

28.  Find  values  of  the  three  angles  of  a  triangle,  if  the 
second  is  three  times  the  first  and  the  third  is  two  times  the 
second. 


34  Plane  Geometry. 

29.  In  a  right  triangle  one  of  the  acute  angles  is  17°  40' 
greater  than  the  other ;  find  the  values  of  both  acute  angles. 

30.  Two  adjacent  angles  are  32°  20'  40"  and  57°  39'  20"; 
what  is  the  relation  of  the  exterior  sides  to  each  other  ? 

31.  Two  adjacent  angles  are  68°  35'  52"  and  111°  24'  8";  what 
is  the  relation  of  the  exterior  sides  to  each  other  ? 

32.  What  part  of  its  complement  is  an  angle  that  is  one 
fourth  of  its  supplement  ?     One  fifth  ?     One  ninth  ? 

33.  What  part  of  its  supplement  is  an  angle  that  is  four 
times  its  complement  ?     Five  times  ?     Six  times  ? 

34.  Four  consecutive  angles  are  formed  at  one  point  on  one 
side  of  a  straight  line ;  three  of  these  angles  are  22°  22'  22" ; 
44°  44'  44" ;  and  55°  55'  55" ;  what  is  the  size  of  the  other  one  ? 

35.  An  exterior  angle  of  a  triangle  is  76°  15'  12",  and  one 
of  the  interior  non-adjacent  angles  is  38°  26'  10";  how  many- 
degrees  in  the  remaining  angles  ? 

36.  The  vertical  angle  of  an  isosceles  triangle  is  35°  45' 55"; 
what  is  the  size  of  the  base  angles  ? 

37.  A  triangle  is  both  isosceles  and  right;  how  many  degrees 
in  each  angle  ? 

38.  How  many  degrees  in  each  angle  of  an  isosceles  triangle 
whose  vertical  angle  is  one  fourth  one  of  the  base  angles? 
One  seventh  ?    One  half  ? 

39.  How  many  degrees  in  each  angle  of  an  isosceles  triangle 
if  one  of  the  base  angles  is  twice  the  vertical  angle  ?  Four 
times  ?    Seven  times  ? 

40.  How  many  degrees  in  each  angle  of  an  isosceles  triangle 
if  the  vertical  angle  is  twice  one  of  the  base  angles  ?  Three 
times  ?    Four  times  ?     Seven  times  ? 


Quadrilaterals.  35 


QUADRILATERALS. 

187.  If  two  lines  in  the  same  plane  be  crossed  by  two  trans- 
versals, a  figure  of  four  sides  may  be  formed,  which  is  called 
a  quadrilateral.  Hence  a  quadrilateral  may  be  defined  as  a 
plane  figure  bounded  by  four  straight  lines.  A  quadrilateral 
is  also  called  a  tetragon. 

188.  If  each  of  the  two  pairs  of  lines  is  parallel,  the  quad- 
rilateral thus  formed  is  called  a  parallelogram. 

Define,  then,  a  parallelogram. 

189.  If  a  parallelogram  have  all  its  sides  equal,  it  is  called 
a  rhombus. 

190.  If  its  angles  are  right  angles,  it  is  called  a  rectangle. 

191.  If  it  is  both  a  rhombus  and  a  rectangle,  it  is  called  a 
square. 

192.  If  a  quadrilateral  have  only  two  of  its  sides  parallel,  it 
is  called  a  trapezoid. 

If  no  two  of  its  sides  are  parallel,  it  is  called  a  trapezium. 

193.  A  parallelogram  whose  angles  are  oblique  and  adjacent 
sides  unequal  is  sometimes  called  a  rhomboid. 

194.  A  rectangle  whose  adjacent  sides  are  unequal  is  some- 
times called  an  oblong. 

195.  The  line  which  joins  two  opposite  vertices  of  a  quad- 
rilateral is  called  a  diagonal. 

196.  The  side  upon  which  a  parallelogram  is  conceived  to 
rest,  and  the  side  opposite  the  latter,  are  termed,  respectively, 
the  lower  and  upper  bases. 

197.  The  parallel  sides  of  a  trapezoid  are  always  considered 
as  its  bases,  the  other  two  sides  its  legs,  while  the  line  bisect- 
ing the  legs  is  called  the  median,  and  the  line  which  bisects 
the  bases  is  called  the  transverse  median. 


36  Plane  Geometry. 

198.  The  altitude  of  either  a  parallelogram  or  trapezoid  is 
the  perpendicular  distance  between  its  bases. 

199.  The  sum  of  the  four  sides  of  a  quadrilateral  is  called 
its  perimeter. 

200.  If  the  legs  of  a  trapezoid  are  equal,  it  is  called  an 
isosceles  trapezoid. 

THEOREMS. 

201.  The  sum  of  the  four  angles  of  a  quadrilateral  equals  four 
right  angles. 

202.  Either  diagonal  divides  the  parallelogram  into  two  equal 
triangles. 

Sug.     Consult  113  and  94. 

203.  The  opposite  sides  of  a  parallelogram  are  equal. 

204.  Converse  of  203. 

Sug.     Draw  one  diagonal  and  consult  95  and  112. 

205.  The  opposite  angles  of  a  parallelogram  are  equal. 

206.  Converse  of  205. 

Sug.     Consult  201  and  119. 

207.  If  two  sides  of  a  quadrilateral  are  equal  and  parallel,  the 
quadrilateral  is  a  parallelogram. 

208.  If  a  quadrilateral  have  two  opposite  sides  equal  and  two 
opposite  angles  equal,  this  quadrilateral  is  a  parallelogram. 

209.  The  two  diagonals  of  a  parallelogram  hisect  each  other. 

210.  Converse  of  209. 

211.  If  one  angle  of  a  parallelogram  be  a  right  angle,  the  other 
^ three  angles    are   right   angles,   and   the 

parallelogram  is  therefore  a  rectangle. 

212.   The  diagonals  of  a  rectangle  are 
equal. 
•^       Sug.     Compare  any   two   triangles 
that  have  one  side  of  the  rectangle  for  a  common  side. 


Quadrilaterals.  3  7 

213.  Converse  of  212. 

214.  The  diagonals  of  a  rhombns  are  perpendicular  to  each  other. 

215.  Converse  of  214. 

216.  The  diagonals  of  a  rhombus  bisect  the  angles. 

217.  If  one  diagonal  of  a  parallelogram  bisects  one  of  its 
angles,  this  parallelogram  is  a  rhombus. 

218.  If  two  parallelograms  have  two  sides  and  their  included 
angle  of  one  equal  respectively  to  two  sides  and  their  included 
angle  of  the  other,  the  two  parallelograms  are  equal  in  every 
respect. 

219.  If  a  line,  parallel  to  the  base  of  a  triangle,  meets  the  other 
two  sides  bisecting  one  of  them,  it  will  also  bisect  the  other  and  be 
equal  to  one  half  the  base.  j^,,;^ -^ 

Sug.     Draw  NS  II  to  AB  and        \^^      ^\7"/^ 
extend  OR  to  meet  it.     Com-  >v  ^^\/r 

pare  the  two  triangles  and  con-  ^v       /     \^ 

sider  the  quadrilateral  AOSN,  \^- ?.>^ 

K  S 

220.  If  a  line  joins  the  middle  points  of  two  sides  of  a  triangle,  it 

wiU  be  parallel  to  the  third  side 
and  equal  to  one  half  of  it. 

221.  The  three  medians  of  a 
triangle  are  concurrent. 

Sug.  In  the  triangle  ABC 
let  CD  and  BF  be  two  medians, 
then  draw  AK  through  H  to 
meet  BK  drawn  II  to  i) (7  and 
join  KC.  Then  consult  219, 
^ .  220,  and  209. 

222.  The  point  of  intersection  of  the  medians  of  a  triangle 
divides  each  median  into  parts  one  of  which  is  twice  as  long  as 
the  other. 

Sug.     Use  previous  diagram. 


38  Plane  Geometry. 

223.  The  line  joining  the  vertex  of  the  right  angle  to  the 
center  of  the  hypothenuse  of  a  right  triangle  is  equal  to  one 
half  the  hypothenuse. 

Sug.  Draw  a  ||  to  one  of  the  legs  from  the  center  of  the 
hypothenuse,  then  consult  219  and  compare  triangles. 

224.  The  median  of  a  trapezoid  is  parallel  to  its  bases  and  equal  to 
one  half  their  sum. 

JSug.  Through  one  extremity  of  the  median  draw  a  line 
parallel  to  one  leg,  and  extend  the  shorter  base  to  meet  it. 

ADVANCE   THEOREMS. 

225.  If  a  quadrilateral  have  one  reflex  angle,  its  exterior  ex- 
plemental  angle  is  equal  to  the  sum  of  the  three  interior 
non-adjacent  angles  of  the  quadrilateral. 

226.  A  line  parallel  to  the  bases  of  a  trapezoid  and  bisecting 
either  diagonal  is  a  median. 

227.  The  line  which  is  parallel  to  the  bases  of  a  trapezoid 
and  bisects  one  leg  is  a  median. 

228.  If  the  angles  at  one  base  of  a  trapezoid  are  equal,  the 
angles  at  the  other  base  are  also  equal. 

229.  If  the  legs  of  a  trapezoid  are  equal,  the  angles  which 
they  make  with  either  base  are  equal,  and  conversely. 

230.  If  the  lesser  base  of  an  isosceles  trapezoid  equals  either 
leg,  the  diagonals  bisect  the  angles  at  the  longer  base. 

231.  If  from  any  point  in  the  base  of  an  isosceles  triangle 
lines  are  drawn  parallel  to  the  equal  sides,  the  perimeter  of  the 
parallelogram  thus  formed  will  be  equal  to  the  sum  of  the  two 
equal  sides  of  the  triangle. 

232.  If  two  opposite  angles  of  a  quadrilateral  are  right 
angles,  the  bisectors  of  the  other  two  angles  are  parallel. 

233.  The  lines  which  join  the  middle  points  of  the  sides  of 
a  triangle  divide  the  triangle  into  four  equal  triangles. 


Theorems.  39 

234.  If  a  line  be  drawn  from  the  middle  point  of  one  side  of 
a  triangle  to  meet  another  side  and  is  equal  to  one  half  the 
base,  it  will  be  parallel  to  the  base. 

235.  If  two  parallel  lines  be  crossed  by  a  transversal,  the 
bisectors  of  the  interior  angles  form  a  rectangle. 

236.  The  lines  which  bisect  the  angles  of  a  rhomboid  form  a 
rectangle. 

237.  The  lines  which  join  the  middle  points  of  the  sides  of 
any  quadrilateral  in  succession  form  a  parallelogram. 

Sug.     Draw  a  diagonal  of  the  quadrilateral,  then  consult  220. 

238.  If  the  middle  points  of  two  opposite  sides  of  a  quadri- 
lateral be  joined  to  the  middle  points  of  the  diagonals,  these 
lines  will  form  a  parallelogram. 

239.  The  lines  which  join  the  middle  points  of  the  sides  of 
a  rhombus  form  a  rectangle. 

240.  The  lines  which  join  the  middle  points  of  the  sides  of 
a  square  form  a  square. 

241.  The  lines  which  join  the  middle  points  of  the  sides  of 
a  rectangle  form  a  rhombus. 

242.  The  lines  which  join  the  middle  points  of  the  sides 
of  an  isosceles  trapezoid  form  a  rhombus. 

243.  The  median  of  a  trapezoid  bisects  both  diagonals. 
Sug.     Consult  224  and  219. 

244.  The  diagonals  of  an  isosceles  trapezoid  are  equal, 
and  conversely. 

245.  If  a  median  of  a  triangle  equals  one  half  the  base, 
this  triangle  is  a  right  triangle. 

246.  If  a  trapezoid  be  isosceles,  the  opposite  angles  are 
supplemental. 

247.  If  a  trapezoid  be  isosceles,  the  angles  which  the 
diagonals  make  with  either  base  are  equal. 


40  Plane  Geometry. 

248.  The  point  of  intersection  of  the  diagonals  of  a  trape- 
zoid bisects  the  line  parallel  to  the  bases  and  terminating 
in  the  legs. 

249.  The  transverse  median  of  a  trapezoid  bisects  any  line 
parallel  to  the  bases  and  terminating  in  the  legs. 

250.  The  line  which  joins  the  middle  points  of  the  diago- 
nals of  a  trapezoid  equals  one  half  the  difference  of  the  bases. 

251.  If  from  the  middle  points  of  two  opposite  sides  of  a 
parallelogram  lines  be  drawn  to  the  vertices  of  the  angles 
opposite,  these  lines  will  trisect  the  diagonal  that  joins  the 
other  two  vertices. 

252.  If  one  of  the  acute  angles  of  a  right  triangle  is  double 
the  other,  the  hypothenuse  is  double  the  shorter  leg. 

253.  If  from  any  two  points  in  the  base  of  an  isosceles 
triangle  perpendiculars  be  drawn  to  the  equal  sides,  the  sum 
of  the  perpendiculars  from  one  point  equals  the  sum  of  the 
perpendiculars  from  the  other  point. 

254.  If  from  any  point  in  an  equilateral  triangle  perpen- 
diculars be  drawn  to  the  sides,  the  sum  of  these  perpendiculars 
is  constant,  and  equal  to  the  altitude  of  the  triangle. 

255.  The  medians  of  any  quadrilateral  bisect  each  other. 

256.  The  point  of  intersection  of  the  medians  of  any  quad- 
rilateral bisects  the  line  which  joins  the  middle  points  of  the 
diagonals. 

257.  The  point  of  intersection  of  the  diagonals  of  a  paral- 
lelogram and  the  middle  points  of  two  opposite  sides  are 
collinear. 

258.  The  point  of  intersection  of  the  diagonals  of  a  paral- 
lelogram bisects  any  line  drawn  through  that  point  and  termi- 
nating in  the  sides. 


Theorems.  41 

259.  If  the  bisectors  of  the  angles  of  a  quadrilateral  form 
another  quadrilateral,  the  opposite  angles  of  the  latter  are 
supplemental. 

260.  If  two  medians  of  a  triangle  are  perpendicular  to  each 
other,  the  third  can  be  made  the  hypoth- 

enuse,  and  the  other  two  the  legs,  of  a   C 
right  triangle.  K^ 

Cons.     Draw  DN  through  F  and  BN      \\\ 

II   t-  ^-  X^K- -^• 

Dem.     DBN is  a  rt.  A.  \^C\ 

AFNB  is  a  parallelogram.  \  /   \\\    /' 

DN=3DF,  and  CK=SHK  jf     KB 
Then  consult  223. 

261.'  If  a  diagonal  of  a  quadrilateral  bisect  two  opposite 
angles,  the  two  diagonals  are  perpendicular  to  each  other. 

262.  If  two  quadrilaterals  be  mutually  equiangular,  and  the 
sides,  including  two  of  the  angles,  respectively  equal,  the  two 
quadrilaterals  are  equal. 


42  Plane  Geometry. 

CIRCLES. 

263.  A  Circle  is  a  portion  of  a  plane  bounded,  by  a  curved 
line,  all  points  of  whicli  are  equidistant  from  a  point  within 
called  the  center. 

264.  The  bounding  line  is  called  the  circumference,  and  any 
portion  of  the  circumference  is  called  an  arc. 

265.  A  Radius  (plural  Radii)  is  any  line  from  center  to 
circumference.  A  Diameter  is  any  line  passing  through  the 
center  and  terminating  both  ways  in  the  circumference. 

266.  A  Semicircumference  is  one  half  the  circumference,  and 
a  Semicircle  is  one  half  the  circle. 

267.  A  Sector  is  that  part  of  a  circle  included  between  an 
arc  and  the  radii  drawn  to  its  extremities,  and  a  Quadrant  is 
a  sector  which  is  one  fourth  the  circle. 

268.  A  Chord  is  any  straight  line  whose  extremities  are  in 
the  circumference. 

269.  A  Segment  is  that  portion  of  the  circle  included  be- 
tween an  arc  and  the  chord  which  joins  its  extremities.  Every 
chord,  therefore,  must  divide  the  circumference  into  two  arcs 
and  the  circle  into  two  segments. 

270.  If  the  arcs  are  unequal,  they  are  designated  as  major  and 
minor  arcs,  and  the  segments  as  major  and  minor  segments. 

271.  The  chord  is  said  to  subtend  the  arc,  and  the  arc  is  said 
to  be  subtended  by  the  chord. 

Whenever  a  chord  and  its  subtended  arc  are  mentioned,  the 
minor  arc  is  meant  unless  it  is  otherwise  specified. 

272.  If  two  circles  have  the  same  center,  they  are  said  to  be 
concentric. 

273.  A  central  angle  is  an  angle  formed  by  two  radii  of  the 
same  circle. 

An  inscribed  angle  is  an  angle  formed  by  two  chords,  with 
its  vertex  in  the  circumference. 


Circles.  43 

274.  An  angle  is  inscribed  in  a  segment  when  its  vertex  is  on 
the  arc  of  that  segment  and  its  sides  meet  the  extremities  of 
the  subtending  chord. 

275.  A  Tangent  to  a  circle  is  a  straight  line  which  has  only 
one  point  in  common  with  the  circumference,  however  far 
extended  both  ways. 

276.  The  point  of  contact  of  a  tangent  to  a  circle  is  the  com- 
mon point  of  the  tangent  and  circumference ;  hence, 

All  points  in  a  tangent  to  a  circle  except  the  point  of  contact 
lie  outside  of  and  beyond  the  circumference. 

277.  Two  circumferences  are  tangent  to  each  other  when 
they  have  one  point  in  common  but  do  not  intersect.  This 
point  is  called  the  point  of  contact,  all  other  points  in  either 
circumference  being  outside  of  and  beyond  the  other. 

278.  If  one  of  two  tangent  circumferences  lies  within  the 
other,  they  are  said  to  be  tangent  internally;  if  it  lies  without, 
they  are  said  to  be  tangent  externally. 

279.  A  Secant  is  a  straight  line  that  intersects  a  circum- 
ference in  two  points  lying  partly  within  and  partly  without 
the  circle;  e.g.  a  chord  extended  in  either  or  both  directions 
becomes  a  secant. 

280.  The  term  circle  is  also  sometimes  used  to  designate  a 
circumference. 

281.  The  distance  of  a  point  from  a  circumference  means  the 
distance  of  that  point  from  the  nearest  point  in  that  circum- 
ference. 

282.  A  triangle  or  quadrilateral  is  inscribed  in  a  circle  when 
the  vertices  of  its  angles  lie  on  the  circumference ;  i.e.  when  its 
sides  are  chords  of  the  circle.  In  this  case  the  circle  is  said  to 
be  circumscribed  about  the  triangle  or  quadrilateral. 


44  Plane  Geometry. 

283.  A  triangle  or  quadrilateral  is  circumscribed  about  a 
circle  when  its  sides  are  tangents  to  the  circle.  In  this  case 
the  circle  is  said  to  be  inscribed  in  the  triangle  or  quadrilateral. 

284.  If  the  circumference  of  a  circle  be  made  to  pass  through 
the  vertices  of  the  angles  of  a  quadrilateral,  this  quadrilateral 
is  said  to  be  inscriptible. 

285.  From  the  foregoing  definitions  the  following  statements 
are  readily  deduced  which  are  axiomatic  in  nature : 

All  radii  of  the  same  circle  are  equal. 

All  diameters  of  the  same  circle  are  equal. 

A  diameter  of  a  circle  is  double  its  radius,  or  the  radius  is 
one  half  the  diameter. 

If  two  circles  be  equal,  their  circumferences  are  equal,  their 
radii  are  equal,  and  their  diameters  are  equal. 

If  two  circles  have  equal  radii,  or  equal  diameters,  or  equal 
circumferences,  the  circles  are  equal. 

If  two  equal  circles  be  concentric,  their  circumferences  will 
coincide. 

Should  there  be  any  doubt  in  regard  to  any  one  of  these,  it 
should  be  promptly  tested  by  superposing  one  circle  upon  the 
other. 

THEOREMS. 

286.  The  point  on  a  circumference  nearest  to  a  given  point  is 
the  point  of  intersection  of  the  circumference  and  hne  joining  the 
given  point  with  the  center  of  the  circle. 

Sug.   There  will  be  two  cases, 
i.   The  given  point  A  is  out- 
side the  circle. 

ii.   The    given    point    B   is 
within  the  circle. 

Then  let  H  and  N  be  any 
points    on    the    circumference 
except  the  points  of  intersection 
j^^^ ^^  S  and  B. 


Circles. 


45 


287.  A  diameter  of  a  circle  is  greater  than  any  other  chord  of  that 
circle. , 

Post   Let  AD  be  a  diameter  and 
BH  any  other  chord. 
jSug.   Consult  89. 

288.  A  diameter  bisects  both  the 
circle  and  its  circumference. 

jSug.  Fold  one  segment  over, 
using  the  diameter  as  an  axis. 

289.  If  a  chord  of  a  circle  bisects 
the  circumference,  this  chord  is  a  diam- 
eter. 

JSug.  Draw  a  diameter  from  one 
end  of  the  chord  and  compare 
arcs.  A 

290.  A  straight  line  cannot  inter- 
sect the  circumference  of  a  circle  in  more 
than  two  points. 

JSug.   Consult  104. 

291.  If,  in  the  same  or  equal  circles,  two  central  angles  be  equal, 
the  arcs  which  their  sides  intercept  will  also  be  equal. 

Sch.  In  this  and  the  following  theorems  where  the  expres- 
sion "  same  or  equal  circles  "  occurs,  the  demonstration  will  be 
more  satisfactory  if  two  circles  are  used. 

Sug.   Apply  one  circle  to  the  other. 

292.  Oonverse  of  291. 

293.  If,  in  the  same  or  equal  circles,  two  chords  be  equal,  the  arcs 
which  those  chords  subtend  are  also  equal. 

Sug.  Draw  radii  to  the  extremities  of  the  chords,  then  con- 
sult 95  and  291. 


294.   Oonverse  of  293. 

Sug.    Draw  radii  as  above. 


Consult  292  and  92. 


46  Plane  Geometry. 

295.  If,  in  the  same  or  equal  circles,  two  central  angles  be  unequal, 
the  arc  intercepted  by  the  sides  of  the  greater  angle  is  greater  than  the 
arc  intercepted  by  the  sides  of  the  lesser  angle. 

jSug.    Take  a  part  of  the  larger  angle  equal  to  the  smaller. 

296.  Converse  of  295. 

297.  If,  in  the  same  or  equal  circles,  two  chords  be  unequal,  the  arc 
subtended  by  the  greater  chord  is  greater  than  that  subtended  by  the 
lesser. 

Sug.    Draw  radii  to  extremities,  then  consult  136  and  295. 

298.  Converse  of  297. 

Sug.  Draw  radii  as  above,  tlien  consult  296  and  135,  or 
three  possible  relations. 

299.  If  a  diameter  be  perpendicular  to  a  chord,  it  will  bisect  the 
chord,  and  also  the  arcs  into  which  the  chord  divides  the  circumference. 

JSug.  Draw  radii  to  extremities  of  the  chord,  then  consult 
134,  291,  and  66. 

Sch.  In  the  subsequent  application  of  this  theorem  it  will 
often  happen  that  only  half  the  diameter  will  be  involved,  in 
which,  case  the  theorem  may  be  quoted  as  follows : 

If  a  radius  be  perpendicular  to  a  chord,  it  will  bisect  the 
chord  and  also  its  subtended  arc. 

300.  A  straight  line  that  is  perpendicular  to  a  radius  at  its 
extremity  is  a  tangent  to  the  circle. 


A  B       jp  a 

Post.    Let  BHKhe  a  circle,  DB  a  radius,  and  AC  a  straight 
line  perpendicular  to  DB  and  passing  through  the  point  B. 


Circles.  47 

To  Prove.    That  AC  is  a  tangent  to  the  circle  BHK. 

Now,  if  we  can  prove  that  every  point  in  AG  except  B  is 
outside  the  circumference,  then  AC  must  be  a  tangent.  (See 
Def.  of  tangent.) 

Cons.    From  D  draw  any  other  line  to  AC,  as  DP. 

Then  P,  the  extremity  of  DP,  must  represent  any  point  in 
AC  except  B. 

Now  DP  >  DB.     Why  ? 

But  DB  is  a  radius,  and  if  DP  is  longer  than  a  radius,  where 
must  its  extremity  be  ? 

Hence,  etc. 

301.  If  a  diameter  or  radius  be  drawn  to  the  point  of  contact  of  a 
tangent  to  a  circle,  it  will  be  perpendicular  to  the  tangent. 

Sug.  Prove  that  the  radius  is  the  shortest  line  from  the 
center  to  the  tangent,  and  use  diagram  similar  to  the  preceding. 

302.  If  a  perpendicular  be  drawn  to  a  tangent  at  the  point  of  con- 
tact, this  perpendicular,  if  extended,  will  pass  through  the  center  of  the 
circle. 

Sug.  Draw  a  radius  to  point  of  contact,  then  consult  301 
and  25. 

303.  If  a  line  be  drawn  from  the  center  of  a  circle  per- 
pendicular to  a  tangent,  this  line  will  pass  through  the  point 
of  contact. 

304.  If  a  chord  and  tangent  be  parallel,  the  arcs  which  they  inter- 
cept are  equal. 

Sug.  Draw  a  radius  to  point  of  contact,  then  consult  301, 
109,  and  299. 

305.  If  two  chords  of  a  circle  be  parallel,  the  arcs  which  they  inter- 
cept are  equal. 

306.  If  two  tangents  of  a  circle  be  parallel,  the  arcs  which  they 
intercept  are  equal. 

307.  The  points  of  contact  of  two  parallel  tangents  and  the 
center  of  the  circle  are  collinear. 


Plane  Geometry. 


308.  If  two  tangents  have  their  points  of  contact  the  ex- 
tremities of  a  diameter,  the  tangents  are  parallel. 

309.  If  a  radius  bisect  a  chord,  it  will  bisect  the  arc  and  be 
perpendicular  to  the  chord. 

310.  If  a  perpendicular  bisect  a  chord,  it  will,  if  extended,  pass 
through  the  center  of  the  circle. 

311.  If  a  radius  bisect  an  arc,  it  will  also  bisect  its  subtend- 
ing chord  and  be  perpendicular  to  it. 

312.  The  middle  points  of  a  chord  and  its  subtended  arc 
and  the  center  of  the  circle  are  collinear. 

313.  The  middle  points  of  two  parallel  chords  and  the  center 
of  the  circle  are  collinear. 

314.  If  two  non-intersecting  chords  intercept  equal  arcs,  they  are 
parallel. 

315.  Converse  of  304. 

316.  If  in  the  same  or  equal  circles  two  chords  be  equal,  they  are 
equidistant  from  the  centers, 

Sug.     Draw  radii  and  compare  the  triangles. 

317.  Converse  of  316. 

318.  If  in  the  same  or  equal  circles  two  chords  be  unequal,  the  lesser 
chord  is  the  farther  from  the  center. 


f> 

/ 

/ 

f 

\^/ 

cV^^ 

S-^' 

Biig.  Make  AB  =  CD.  Draw  HS  through  B  and  HN  to 
bisect  AR.  Compare  angles  A  and  C  by  means  of  A  AHK  and 
CPQ  (see  126,  296,  etc.),  then  compare  JT^and  PQ  (see  135). 


Circles. 


49 


319.   Converse  of  318. 

Sug.    Three  possible  relations,  or 


Draw  chord  BF^NR,  QTTXto  BF,  and  join  KX-,  then 
consult  107  and  97. 

320.  Through  any  three  pomts  that  are  not  in  the  same  straight 
line,  one  circumference  can  be  made  to  pass,  and  only  one;  or,  if 
more  be  drawn,  they  must  coincide. 

Sug.    Join  two  pairs  of  points,  then  consult  100. 

321.  If  two  unequal  circles  be  concentric,  the  chords  of  the 
greater,  which  are  tangents  of  the  less,  are  equal. 

Sug.     Consult  317. 

322.  Two  unequal  circles  may  have  five  positions  relative  to  each 
other,  viz. : 

i.   One  may  lie  wholly  within  the  other  without  contact. 
ii.   They  may  be  tangent  to  each  other  internally. 


Case  I. 


Case  II. 


50 


Plane  Geometry. 


iii.  Their  circumferences  may  intersect  each  other. 


Case  III. 
iv.   They  may  be  tangent  to  each  other  externally. 


Case  IV. 
V.   One  may  lie  wholly  without  the  other  without  contact. 


Case  V. 


Circles.  5 1 


AB  is  the  line  joining  the  centers.  Designating  the  radius 
of  the  larger  circle  by  H  and  radius  of  the  smaller  by  r,  the 
pupil  should  ascertain  the  relation  between  AB  and  M  +  r  and 
AB  and  B  —  r  in  each  case. 

What  must  be  the  position  of  two  circles,  then,  if  the  dis- 
tance between  their  centers  is 
i.  0? 

ii.   Less  than  the  difference  of  their  radii  ? 

iii.  Equal  to  the  difference  of  their  radii  ? 

iv.  Less  than  the  sum,  and  greater  than  the  difference,  of 
their  radii  ? 

V.   Equal  to  the  sum  of  their  radii  ? 

vi.   Greater  than  the  sum  of  their  radii  ? 

323.  If  two  circles  be  tangent  to  each  other  externally,  the  point 
of  contact  and  the  center  of  the  two  circles  are  collinear. 

Post,  Let  the  two  circles  PCN  and  CDK  be  tangent  to 
each  other  externally,  the  point  of  contact  being  C,  and  let  A 
and  B  be  their  respective  centers. 


To  Prove.  A^  C,  and  B  are  in  the  same  straight  line,  i.e. 
are  collinear. 

Cons.  Let  the  radii  AC  and  BC  be  drawn,  and  BD  any 
other  radius  of  the  circle  DCK,  and  points  A  and  D  joined. 

If  we  can  prove  that  ACB  is  the  shortest  line  between  A 
and  B,  then  it  is  a  straight  line,  and  the  three  points  are 
collinear. 


52  Plane  Geometry. 

So  we  are  to  prove  that  ACB  is  the  shortest  distance  from 
AtoB. 

Dem,  Since  every  point  in  the  circumference  CDK  except 
C  is  outside  the  circumference  CPJ!^  (see  Def.),  AD  must 
extend  beyond  the  circumference. 

AD>AH.     Why?  DB=CB.  Why? 

AC=AH.    Why ?  .-.  AD  +  DB>AG^  CB.    Why ? 

.-.  AD>Aa     Why? 

But  D  is  any  point  in  the  circumference  CDK  except  C. 
Hence,  the  distance  from  JL  to  ^  by  way  of  D  is  always 
greater  than  by  way  of  C.  Or  ACB  is  the  shortest  distance 
between  A  and  B,  and  is  therefore  a  straight  line. 

.-.  A,  B,  and  C  are  coUinear  since  they  are  in  the  same 
straight  line.  q.e.d. 

324.  If  two  circles  are  tangent  to  each  other  internally, 
their  centers  and  point  of  contact  are  collinear. 

325.  If  the  circumferences  of  two  circles  intersect  each 
other,  the  line  joining  their  centers  will  bisect  their  com- 
mon chord  and  be  perpendicular  to  it. 

Sug.     Consult  103  or  95  and  141. 

326.  If  the  circumferences  of  two  equal  circles  intersect 
each  other,  their  common  chord  will  bisect  the  line  joining 
their  centers. 

327.  If  from  the  same  point  without  a  circle  two  tangents 
be  drawD,  these  two  tangents  are  equal ;  i.e.  the  distances  from 
the  common  point  to  the  points  of  contact  are  equal. 

JSug.    Consult  134. 

328.  If  from  a  point  in  a  secant  that  passes  through  the 
center  of  the  circle  two  tangents  be  drawn,  the  secant  will 
bisect  the  angle  formed  by  the  two  tangents,  the  chord  joining 
the  points  of  contact,  and  its  subtended  arc. 


Measurement.  53 


MEASUREMENT,  RATIO,   AND  PROPORTION. 

329.  To  measure  a  quantity  of  any  kind  is  to  find  how  many 
times  it  contains  another  quantity  of  the  same  kind  used  as 
the  unit;  thus,  to  measure  a  line  is  to  find  the  number  which 
expresses  how  many  times  it  contains  another  line  called  the 
unit  of  length,  or  liiiear  unit,  as  the  foot,  yard,  rod,  etc.  This 
number  is  called  the  numerical  measure  of  that  quantity. 

330.  If  of  two  unequal  magnitudes  the  less  is  contained  an 
exact  number  of  times  in  the  greater,  the  latter  is  said  to  be  a 
multiple  of  the  former,  and  the  former  an  aliquot  part  of  the 
latter. 

331.  If  of  three  unequal  magnitudes  the  smallest  is  con- 
tained an  exact  number  of  times  in  each  of  the  two  larger,  the 
latter  are  said  to  be  commensurable,  and  the  former  is  said  to 
be  their  common  measure. 

332.  When  no  magnitude  can  be  found  which  is  contained 
an  exact  number  of  times  in  each  of  the  two  larger,  the  latter 
are  said  to  be  incommensurable. 

333.  If  two  commensurable  magnitudes  contain  their  com- 
mon measure,  one  n  times  and  the  other  m  times,  they  are  said 
to  be  to  each  other  as    ^ ^ 

n  to  m,  or  in  the  ratio    b F 

of  n  to  m.  ^ '——— — ^ 

For  example,  if  the  line  AB  is  contained  in  the  line  DF  3 
times,  and  in  line  NR  5  times,  then  DF  is  to  NR  as  3  to  5, 
and  the  line  AB  is  a  common  measure  of  the  two  lines  DjF' and  NR. 

334.  Equimultiples  of  two  or  more  magnitudes  are  the  results 
obtained  by  multiplying  these  magnitudes  by  the  same  number 

(see  46). 


Thus,  if  AB  and  NO  be 


A B 

J) jj< 

^ ^   any  two  lines,  and  DF  and 

iV 0  HK  two  other  lines  such  that 


54  Plane  Geometry. 

the  former  contains  the  line  AB  n  times  (in  this  particular 
case  n  is  3),  and  the  latter  contains  the  line  NO  n  times,  then 
DF  and  HK  are  equimultiples  of  AB  and  NO  respectively. 

335.  If  two  magnitudes  are  to  each  other  as  m  to  w  (see  333), 
it  is  usually  expressed  thus,  m :  n,  called  the  ratio  form ;  or  thus, 

— ,  called  the  fractional  form. 

336.  A  Ratio  may  be  defined  as  an  expression  of  comparison, 
in  respect  to  size,  of  two  magnitudes  of  the  same  kind. 

337.  If  AB  and  DF  are  two  magnitudes,  and  DF  a  multiple 

^ B  of  AB  (see  330),  the  latter  being 

J) y    contained   in  the   former  n  times, 

then  the  ratio  of  DF  to  AB  is  as  n  to  unity,  or  w  ;  1. 

338.  If  X  and  y  are  commensurable  (331),  and  their  common 
measure  is  contained  n  times  in  a;,  and  m  times  in  2/,  then  the 
ratio  of  a;  to  2/  is  as  n  to  m,  or,  technically  expressed,  n :  m. 

339.  If  X  and  y  are  incommensurable  (see  332),  the  ratio  can- 
not be  exactly  expressed.  If,  however,  x  and  y  are  numbers, 
the  ratio  may  be  approximately  expressed  by  placing  1  for  the 
first  term,  and  the  quotient  of  the  greater  divided  by  the  less 
to  any  number  of  decimal  places  for  the  second  term. 

340.  The  first  term  of  a  ratio  is  called  the  antecedent^  and 
the  second  term  the  consequent. 

341.  A  Proportion  is  an  expression  of  equality  between  two 
equal  ratios,  usually  indicated  by  four  dots  between  the  two 
ratios ;  thus,  aibiixiy,  and  read,  a  is  to  6  as  a;  is  to  y,  or 

-  =  -  called  the  equation  form. 
b     y 

342.  The  four  magnitudes  forming  a  proportion  are  called 
proportionals. 

343.  If  a  proportion  contains  only  two  ratios,  it  is  called  a 
simple  proportion ;  if  more  than  two  and  all  equal,  it  is  called 
a  continued  proportion. 


Ratio  and  Proportion.  55 

344.  The  first  and  last  terms  of  a  simple  proportion  are 
called  the  extremes,  and  the  other  two  the  means. 

345.  In  a  simple  proportion,  where  the  terms  are  all  of 
different  values,  each  term  is  said  to  be  a  fourth  proportional 
to  the  other  three. 

346.  In  a  simple  proportion,  where  the  two  means  or  the 
two  extremes  are  alike,  the  repeated  quantity  is  said  to  be  a 
mean  proportional  to  the  other  two,  and  the  proportion  is 
called  a  mean  proportion  ;  also,  either  of  the  quantities  not 
repeated  is  said  to  be  a  tliird  proportional  to  the  other  two. 

347.  When  a  line  is  divided  by  a  point  between  its  ex- 
tremities, it  is  said  to  be  divided  into  segments  internally; 
when  the  point  of  division  is  on 

the  extension  of  the  line,  it  is  said  A • — B 

to  be  divided  externally.     Thus,  AB    N ^— O 

is  divided  internally  at  F,  and  PO 

is  divided  externally  at  N.    The  segments  in  both  cases  are  the 

distances  between  the  point  of  division  and  the  ends  of  the  line. 

348.  When  a  line  is  so  divided  that  the  larger  part  is  a  mean 
proportional  between  the  whole  line  and  the  smaller  part,  it  is 
said  to  be  divided  in  extreme  and  mean  ratio;  and  when  it  is 
divided  both  internally  and  externally  into  segments  having 
the  same  ratio,  it  is  said  to  be  divided  harmonically, 

349.  It  will  be  found  on  investigation  that  the  order  of  four 
proportionals,  when  all  of  the  same  kind,  can  be  varied,  as  well 
as  certain  other  transformations  effected,  without  destroying 
the  proportion.     The  principal  of  these  are  as  follows : 

350.  Magnitudes  are  said  to  be  in  proportion  by  alternation 
when  either  the  two  means  or  the  two  extremes  are  made  to 
exchange  places. 

351.  Magnitudes  are  said  to  be  in  proportion  by  inversion 
when  the  means  are  made  to  exchange  places  with  the 
extremes. 


56  Plane  Geometry. 

352.  Magnitudes  are  said  to  be  in  proportion  by  composition 
when  the  sum  of  the  terms  of  the  first  ratio  is  to  either  term  of 
the  first  ratio  as  the  sum  of  the  terms  of  the  second  ratio  is  to 
the  corresponding  term  of  that  ratio. 

353.  Magnitudes  are  said  to  be  in  proportion  by  division 
when  the  difference  of  the  terms  of  the  first  ratio  is  to  either 
term  of  that  ratio  as  the  like  difference  of  the  terms  of  the 
second  ratio  is  to  the  corresponding  term  of  that  ratio. 

354.  Magnitudes  are  said  to  be  in  proportion  by  composition 
and  division  when  the  sum  of  the  terms  of  the  first  ratio  is  to 
their  difference  as  the  sum  of  the  terms  of  the  last  ratio  is  to 
their  like  difference. 

THEOREMS. 

355.  In  the  following  theorems  the  "  product  of  two  quanti- 
ties "  or  the  "  product  of  two  magnitudes  '^  means  the  product 
of  the  numbers  representing  their  numerical  measurement.  It 
is  assumed  that  pupils  are  familiar  with  elementary  algebraic 
processes. 

356.  If  four  magnitudes  are  proportionals,  the  product  of  the  means 
equals  the  product  of  the  extremes. 

Given  a:x::n:7\ 


To  Prove 

ar  =  nx. 

Dem. 

a:x::n:r. 

(By  Hyp.) 

a_n^ 
x"  r 

(Changing 

to  equation 
form.) 

nx 
,'.  a  —  — . 
r 

(Ax.  viii.) 

.-.  ar  =  nx. 

(Ax.  viii.) 

Q.E.D. 

^ch.  The  fact  stated  in  the  above  theorem  is  the  test  of  a 
proportion,  and  may  be  applied  to  settle  any  doubt  or  question 
regarding  it. 


Ratio  and  Proportion.  57 

357.  If  the  product  of  two  quantities  equals  the  product  of  two 
others,  the  factors  of  either  product  may  be  made  the  means,  and  the 
other  two  the  extremes,  of  a  proportion. 

Post.     Let  ex  =  an. 

We  are  required  to  form  a  proportion  from  the  four  factors 
c,  X,  a,  and  n,  in  which  c  and  x  shall  be  the  extremes. 


Dem.  ex  =  an. 


"a     x' 

(?)  (?) 

an 
.*.  c  =  — .  .'.  c:a::n:x, 

(?)  *  (?) 

Q.E.D. 

Take  the  same  equation  and  form  other  proportions,  i.e.  make 
the  terms  appear  in  a  different  order. 

Form  proportions  from  the  following  equations  in  which  the 
factors  of  the  product  marked  Ex.  shall  form  the  extremes, 
avoiding  the  factor  1. 

Ex. 

i.  2x  —  ajc. 


ii. 

Ex. 

6  =  aa;. 

iii. 

Ex. 

3Va  =  14. 

iv. 

Ex. 

a{x-\-y)  =  n\ 

V. 

Ex. 

'if  =  an-^ac. 

vi. 

Ex. 

ah  -\- hx  =  en  -\-  ye. 

vii. 

Ex. 

ax  -{-x  =  yh^  -\-  W. 

dii. 

Ex. 

a?-l^a^-h\ 

ix. 

Ex. 

ax  -\-  xb  -\-  ex  =  nd  -\-  7in  -\-  nJc. 

X. 

Ex. 

ai^  -\- 2  ex -{- c^  =  an  -{- ny. 

58  Plane  Geometry. 

358.  If  fonr  qnantities  form  a  proportion,  they  will  be  in  propor- 
tion by  Inversion.        JSug.    Consult  357. 

359.  If  four  quantities  form  a  proportion,  they  will  be  in  propor- 
tion by  Alternation.        Sug.     Consult  357. 

360.  If  four  quantities  form  a  proportion,  they  will  be  in  pro- 
portion by  Composition. 

Post.    Let  the  four  quantities  x,  y,  a,  and  b  form  a  propor- 
tion, so  that  x:y::a:b. 
To  Prove.            1.   x-\-y:y::a-{-b:b. 
ii.   x-}-y:x::a-{-b:a. 
Dem, 


x:y::a:b. 

(By  Hyp.) 

X     a 

Why? 

^  +  1  =  ^  +  1. 

y             b 

Why? 

y     y     b     b 

Why? 

x-{-y     a-\-b 
y          b 

Why? 

x-{-y:y::a-\-b:b. 

Why? 

Q.E.D. 

The  pupil  should  demonstrate  Part  ii. 

361.  If  four  quantities  form  a  proportion,  they  will  be  in  pro- 
portion by  Division. 

JSug.  There  are  four  cases.  In  Case  i.  subtract  1  from  each 
member  of  the  equation.  The  other  cases  may  be  similarly 
demonstrated  by  consulting  358.  » 

362.  If  two  proportions  have  a  ratio  in  each  equal,  the  other  two 
ratios  will  form  a  proportion. 

363.  If  two  proportions  have  the  two  antecedents  of  one  equal 
respectively  to  the  two  antecedents  of  the  other,  the  consequents  will 
form  a  proportion. 


Ratio  and  Proportion.  59 

364.  If  two  proportions  have  the  two  consequents  of  one  equal 
respectively  to  the  two  consequents  of  the  other,  the  antecedents  will 
form  a  proportion. 

365.  If  the  terms  of  one  ratio  of  a  proportion  are  equal,  the  terms 
of  the  other  ratio  or  ratios  are  also  equal. 

366.  If  four  quantities  form  a  proportion,  they  will  be  in  propor- 
tion by  Composition  and  Division. 

Sug.     Consult  360  and  301,  and  363  or  364. 

367.  If  any  number  of  magnitudes  of  the  same  kind  form  a  pro- 
portion, the  sum  of  the  antecedents  is  to  the  sum  of  the  consequents 
as  any  antecedent  is  to  its  consequent. 

Post.     Let  the  quantities  a,  ic,  c,  n,  d,  and  r  form  a  continued 

proportion,  so  that 

a  :  X : :  c  ',  n  : :  d  '.  r. 

To  Prove.        i.   a  +  c  +  cZ:a;  +  w-|-r::a:a;; 
or  ii.  : :  c :  71 ; 

or  iii.  i:d:  r. 

Dem.  a:x::c:n::d:r.  (By  Hyp.) 

.-.  (1)  ax  =  ax'j  Why? 

(2)  cx  =  an;  Why? 

(3)  dx=:ar.  Why? 
Hence,              ax  •{- ex -\- dx  =  ax -{■  an -\-  ar,       Why  ? 

or  a;  (a  -f-  c  +  c?)  =  a  (a;  +  n  +  r).        Why  ? 

The  pupil  should  be  able  to  finish  this  case,  and  also  dem- 
onstrate the  other  two  without  diJfficulty.     Consult  357. 

368.  If  four  quantities  form  a  proportion,  the  terms  of 
either  ratio  may  be  either  multiplied  or  divided  by  the  same 
quantity,  and  the  results  still  form  a  proportion. 

369.  If  the  antecedents  or  consequents  of  a  proportion  be 
either  multiplied  or  divided  by  the  same  quantity,  the  results 
will  still  form  a  proportion. 


6o  Plane  Geometry. 

370.  If  four  quantities  form  a  proportion,  and  the  terms  of 
one  ratio  be  either  multiplied  or  divided  by  the  same  quantity, 
while  both  terms  of  the  other  ratio  be  either  multiplied  or 
divided,  either  by  the  same  or  different  quantity  from  that 
used  in  the  first  ratio,  the  results  will  still  form  a  proportion. 

Sug.  The  pupil  should  take  each  case  involved  in  the  state- 
ment of  the  above  theorem  separately. 

371.  If  four  magnitudes  form  a  proportion,  and  the  ante- 
cedents be  either  multiplied  or  divided  by  the  same  quantity, 
while  the  consequents  are  both  multiplied  or  divided  by  another 
quantity,  the  results  will  form  a  proportion. 

372.  If  two  proportions  be  "given,  the  products  of  the  correspond- 
ing terms  will  also  form  a  proportion. 

373.  If  two  proportions  be  given  in  which  two  correspond- 
ing ratios  have  the  antecedent  of  one  equal  to  the  consequent 
of  the  other,  the  remaining  antecedent  and  consequent,  together 
with  the  products  of  the  corresponding  terms  of  the  other  two 
ratios,  will  form  a  proportion. 

374.  If  the  antecedents  of  a  proportion  are  equal,  the  con- 
sequents are  equal. 

375.  Converse  of  37'4. 

376.  If  four  quantities  form  a  proportion,  the  reciprocals  of 
the  terms  of  either  ratio  are  inversely  proportional  to  the  terms 
of  the  other  ratio. 

377.  If  four  quantities  form  a  proportion,  their  like  powers  and 
like  roots  will  also  form  a  proportion. 

378.  If  three  quantities  form  a  proportion,  the  first  is  to  the 
third  as  the  square  of  the  first  is  to  the  square  of  the  second ; 
i.e.  if  a  :  n  : :  n  :  X  .'.  a  :  X '.  :  o?  :  n^. 

379.  If  three  terms  of  a  simple  proportion  are  equal  respectively 
to  the  three  corresponding  terms  of  another  proportion,  the  fourth 
terms  of  the  two  proportions  are  equal. 

380.  Equimultiples  of  two  magnitudes  are  in  the  same  ratio  as 
the  magnitudes  themselves. 


Theory  of  Limits.  6i 


THEORY   OP  LIMITS. 

381.  A  Constant  Quantity,  or  simply  a  Constant^  is  a  quantity 
whose  value  remains  unchanged  throughout  the  same  discussion. 

382.  A  Variable  Quantity,  or  simply  a  Variable,  is  a  quantity 
which  may  assume  different  values  in  the  same  discussion, 
according  to  the  conditions  imposed. 

383.  The  Limit  of  a  variable  is  a  constant  quantity,  which 
the  variable  is  said  to  approach  in  value  whenever  a  regular 
and  definite  increase  or  decrease  in  value  is  assigned  to  the 
latter. 

384.  \ATienever  it  can  be  shown  that  the  value  of  a  variable, 
by  such  constant  increase  or  decrease  in  value,  can  be  made  to 
differ  from  that  of  its  limit  by  less  than  any  appreciable  or 
assignable  quantity,  however  small,  this  variable  is  said  to 
approach  indefinitely  to  its  limit. 

For  example : 

*  -     -  t 

A  C  D  H  KB 

Suppose  a  point  move  from  A  toward  B  under  the  condition 
that  during  the  first  second  it  shall  move  over  one  half  the 
distance  AB,  or  AC,  and  that  during  each  successive  second 
it  shall  move  over  one  half  the  remaining  distance.  Then  at 
the  end  of  the  second  second  it  would  be  at  D,  at  the  end  of 
the  third  at  H,  at  the  end  of  the  fourth  at  K,  and  so  on.  It 
is  evident  that  it  can  never  reach  the  point  B,  for  there  will 
constantly  remain  one  half  the  distance ;  but  if  its  motion  be 
continued  indefinitely,  it  will  approach  indefinitely  near  to  B. 

Consequently,  the  distance  from  A  to  the  moving  point  is 
an  increasing  variable  and  AB  is  its  limit ;  while  the  distance 
from  B  to  the  moving  point  is  a  decreasing  variable,  with  zero 
as  its  limit. 

Other  illustrations  may  be  given,  e.g. : 

0.3333  +  ...  =  A  +  ^^  +  ^^  +  .r^,  +  .... 


62  Plane  Geometry. 

Here  the  sum  of  the  series  of  fractions  is  the  increasing 
variable,  and  approaches  ^  as  its  limit. 

Again:  let  ABC  be  a  right  triangle,  with 
C  the  right  angle,  and  consider  the  point  B  to 
move  toward  (7;  the  angle  A  will  then  be  a 
decreasing  valuable  approaching  zero  as  its  limit, 
and  the  angle  B  will  be  an  increasing  variable 
approaching  a  right  angle  as  its  limit. 

385.  It  will  be  evident  to  the  pupil  that  if  the  numerator  of 
a  fraction  be  a  variable  and  the  denominator  a  constant,  then 
the  entire  fraction  may  be  regarded  as  a  variable,  increasing 
or  decreasing  as  the  numerator  is  an  increasing  or  decreas- 
ing variable.  Similarly,  if  the  numerator  be  a  constant  and 
the  denominator  a  variable,  the  entire  fraction  is  a  variable, 
increasing  or  decreasing  according  as  the  denominator  is  a 
decreasing  or  increasing  variable. 

386.  Theorem.  If  two  variables  are  always  equal,  and  each 
approaches  a  limit,  their  hmits  are  equal. 

A — ^ J3 


F- 


N 


-B 


Post.  Let  AB  and  FE  be  the  limits  to  which  the  two  equal 
variables  AD  and  FN  indefinitely  approach. 

To  Prove.  AB  =  FB. 

Dem.  AB  and  FR  are  either  equal  or  unequal.  Let  us 
suppose  them  unequal,  and  that  AB  is  the  greater.  Mark  off 
AQ  equal  to  FR. 

Then  the  variable  i<W  cannot  exceed  FR,  but  the  variable  AD 
may  exceed  AQ,  and  consequently  the  variable  AD  becomes 
greater  than  the  variable  FN.  This,  however,  is  contrary  to 
the  hypothesis  that  the  two  variables  must  always  be  equal. 
Therefore,  AB  and  FR  cannot  be  unequal ;  i.e.  they  are  equal. 

Q.E.D. 


Theory  of  Limits.  63 

387.   In  the  same  or  equal  circles  two  central  angles  are  in  the 
same  ratio  as  the  arcs  which  their  sides  intercept. 


Post.  Let  NBA  and  QKH  be  two  equal  circles,  and  C  and 
D  two  central  angles. 

To  Prcyve.     Z  C:  Z  D:  :3i.TcAB:BJ:o  HK. 

'    Case  I.     When  the  angles  are  commensurable, 

Dem.  If  the  angles  are  commensurable,  there  is  some  angle, 
as  Wy  which  will  be  contained  an  exact  number  of  times  in  each. 

Suppose  it  is  contained  m  times  in  Z  O  and  n  times  in  Z  D. 

Then  ZC :  ZD:'.m:n. 

If,  now,  lines  be  drawn  from  G  and  D  dividing  the  two 

angles  into  m  and  n,  equal  parts  respectively,  each  part  being 

equal  to  angle  W,  then  arc  AB  will  be  divided  into  m  equal 

arcs,  and  HK  into  n  equal  arcs,  the  divisions  all  being  equal, 

by  Theorem  291. 

.*.  arc  AB  :  arc  ^/f: :  m  :  w. 

.-.  ZC\ZD::qxqAB:qiq,HK. 

(?) 
Case  II.     When  the  angles  are  incommensurable. 
Post.     Let  AQB  and  NHK  be  two  equal  circles,  and  C  and 
D  two  central  angles  which  are  incommensurable. 

To  Prove.  ZD^arcHK 

ZO      arcAB' 

or  ZD:ZC::SiicHK:siTcAB. 


64  Plane  Geometry. 


Dem.  Conceive  the  Z  (7  to  be  divided  into  any  number  of 
equal  parts,  and  one  of  these  parts  to  be  applied  as  a  unit 
of  measure  to  the  /.D.  It  will  be  contained  a  certain  num- 
ber of  times  with  a  remainder,  SDK,  less  than  the  unit  of 
measure. 

If,  now,  the  number  of  equal  parts  into  which  the  Z  (7  is 
divided  be  indefinitely  increased,  the  unit  of  measure  of  the 
Z  HDK  will  be  correspondingly  diminished,  and  the  point  S 
will  get  indefinitely  near  to  K. 

.'.  Z  HDS  becomes  an  incr.  var.  appr.  its  limit  Z  HDK. 

.-.  —  becomes  an  incr.  var.  appr.  its  limit  — 777-* 

Z  C  Z  (7 

In  like  manner, 

arc  HS  becomes  an  incr.  var.  appr.  its  limit  arc  HK 

...  arc becomes  an  incr.  var.  appr.  its  limit  ^^ — - — 

arc  AB  arc  AB 

But  ZHDS^^jcHS (.^3^  J 

ZO        arc  ^5 

ZHDK  ^BxcHK  ,QogN 

or  Z  D :  Z  O : :  arc  HK:  arc  AB,  q.e.d. 


Circles.  65 

388.    If  in  the  same  or  equal  circles  one  central  angle  be  used  as 
a  unit  to  measure  another,  and  its  inter- 
cepted arc,  as  a  unit  to  measure  tlie  other 
arc,  then  their  numerical  measures   are 
equal. 

Post.  Let  the  central  angle  ABR 
be  used  as  a  unit  to  measure  the 
central  angle  ABC,  and  arc  AR  as 
a  unit  to  measure  the  arc  AG. 

To  Prove.  That  their  numerical 
measures  are  equal. 

Dem.      Z  ABC :  Z  ABR  : :  arc  AC :  arc  AR     (Theorem  387) 

Z.ABC     Sivc  AC 


or 


ZABR     2.YQAR 


That  is,  the  angle  ABR  is  contained  in  the  angle  ABC  the  same 
number  of  times  that  the  arc  AR  is  contained  in  the  arc  AC. 
Hence  their  numerical  measures  are  equal.  q.e.d. 

389.  Sch.  If  the  degree  (see  185)  were  used  to  measure 
the  angle,  and  its  corresponding  arc  to  measure  the  arc, 
then,  from  the  above  demonstration,  there  would  be  the 
same  number  of  angle  degrees  as  arc  degrees;  and  so  the 
arc  may  be  said  to  measure  the  angle.  Formal  expression  to 
this  idea  is  commonly  given  as  follows,  viz. :  A  central  angle 
is  meas^ired  by  the  arc  which  its  sides  intercept  on  the  circumfer- 
ence. Hence,  whatever  numerical  ratio  exists  between  two 
arcs  in  the  same  or  equal  circles,  the  same  numerical  ratio 
exists  between  the  central  angles  whose  sides  intercept  those 
arcs;  e.g.  if  in  the  same  or  equal  circles  two  central  angles 
intercept  arcs  one  of  which,  is  twice  as  large  as  the  other,  then 
one  angle  is  twice  as  large  as  the  other,  etc.  Similarly,  if  an 
angle  is  measured  by  one  fourth  the  circumference,  this  angle 
is  a  right  angle ;  and  if  two  angles  are  together  measured  by 
one  half  the  circumference,  these  angles  are  supplemental,  etc. 


66 


Plane  Geometry. 


390.  If  in  the  same  or  equal  circles,  an  inscribed  angle  and  a  cen- 
tral angle  intercept  the  same  or  equal  arcs,  the  central  angle  is  double 
the  inscribed  angle. 

B  K 


A     o 

JSug.   Consult  292  and  129. 

391.  An  inscribed  angle  is  measured  by  one  half  the  arc  intercepted 
by  its  sides.  In  demonstrating  this  theorem  it  will  simplify  it  some- 
what to  make  three  cases  of  it ;  viz : 

I.  When  one  of  the  sides  of  the  angle  is  a  diameter. 
II.   When  the  center  is  between  the  sides  of  the  angle. 
III.  When  the  center  is  without  the  angle. 
Post. 
To  Prove. 
Cons.  Draw  a  diameter,  as  HK,  parallel  to  BD. 

Dem.  What  relation  exists  be- 
tween the  angles  BCK  and  HCA? 
Why? 

What  relation,  then,  exists  between 
the  two  arcs  HA  and  BK?    Why  ? 

What  relation  between  the  two  arcs 
DHsLTidBK?    Why? 

What  relation,  then,  between  the 
two  arcs  DH  and  AH  ?    Why  ? 

What  relation,   then,  between  the 
two  arcs  AH  and  AD  ?    Why  ? 
What  relation  between  the  two  angles  HCA  and  B  ?    Why  ? 


Circles.  .  67 

What  measures  the  angle  HGA  ?    Why  ? 

What,  then,  must  measure  the  angle  B  ?    Why  ? 

After  answering  correctly  the  foregoing  questions  the  pupil 
should  have  no  difficulty  in  writing  out  a  complete  demonstra- 
tion of  each  of  the  three  cases,  using  Case  I  in  demonstrating 
II,  etc. 

392.  If  two  angles  be  inscribed  in  the  same  or  equal  circles,  and 
their  sides  intercept  the  same  or  equal  arcs,  the  two  angles  are  equal. 

393.  Converse  of  392. 

394.  An  angle  inscribed  in  a  semi- 
circle is  a  right  angle. 

Sug.  Tiove  ^  ABC  =  ZCBF. 

395.  An  angle  inscribed  in  a 
segment  greater  than  a  semicircle 
is  an  acute  angle,  and  an  angle 
inscribed  in  a  segment  smaller 
than  a  semicircle  is  an  obtuse  angle. 

396.  If  an  angle  be  formed  by  two  chords  whose  vertex  is  between 
the  center  and  circumference,  it  will  be  measured  by  one  half  the  sum 
of  the  arcs  intercepted  by  its  sides  and  the  sides  of  its  vertical  angle. 

397.  If  the  vertex  of  an  angle  formed  by  two  secants  is  without 
the  circle,  this  angle  is  measured  by  one  half  the  difference  of  the  two 
intercepted  arcs. 

398.  If  an  angle  be  formed  by  a  tangent  and  chord,  this  angle  is 
measured  by  one  half  the  arc  intercepted  by  its  sides. 

399.  If  the  vertex  of  an  angle  formed  by  a  tangent  and  secant  is 
without  the  circle,  this  angle  is  measured  by  one  half  the  difference  of 
the  two  intercepted  arcs. 

400.  The  angle  formed  by  two  tangents  to  the  same  circle  is  meas- 
ured by  one  half  the  difference  of  the  two  intercepted  arcs. 


C 


68  Plane  Geometry. 

401.  The  opposite  angles  of  an  inscribed  quadrilateral  are 
supplemental. 

402.  Sch.  The  principle  enunciated  in  389  with  reference  to 
central  angles  may  now  be  stated  with  reference  to  any  two 
angles  relating  to  the  same  or  equal  circles ;  viz : 

If,  in  the  same  or  equal  circles,  two  angles  are  known  to  be 
in  a  certain  numerical  ratio,  their  measuring  arcs  will  be  in  the 
same  ratio,  and  conversely. 

ADVANCE  THEOREMS. 

403.  If,  from  the  same  point,  two  tangents  to  a  circle  be 
drawn  whose  points  of  contact  are  the  extremities  of  a  chord, 
the  angle  formed  by  the  two  tangents  is  double  the  angle 
formed  by  the  chord  and  diameter  drawn  from  either  extremity 
of  the  chord. 

404.  The  bisectors  of  the  angles  of  a  circumscribed  quadri- 
lateral are  concurrent. 

405.  If  two  circles  which  are  tangent  to  each  other  exter- 
nally have  three  common  tangents,  the  one  that  passes  through 
the  point  of  contact  of  the  two  circles  bisects  the  other  two. 

JSug.  'Consult  327. 

406.  If  four  tangents  form  a  quadrilateral,  the  sum  of  two 
opposite  sides  equals  the  sum  of  the  other  two  opposite  sides. 

407.  If  two  mutually  equiangular  triangles  be  inscribed  in 
equal  circles,  the  triangles  are  equal  in  every  respect. 

408.  If  two  opposite  sides  of  an  inscribed  quadrilateral  are 
equal,  the  other  two  sides  are  parallel. 

409.  Converse  of  408. 

410.  If  the  opposite  angles  of  a  quadrilateral  are  supple- 
mental, it  is  inscriptible. 

411.  If  two  equal  chords  be  drawn  from  opposite  extremities 
of  a  diameter  and  on  opposite  sides  of  it,  they  will  be  parallel. 


Circles.  69 

412.  If  two  chords  be  drawn  through  the  same  point  in  a 
diameter  making  equal  angles  with  it,  thej  are  equal. 

413.  If  one  of  the  equal  sides  of  an  isosceles  triangle  be  the 
diameter  of  a  circle,  the  circumference  will  bisect  the  base. 

414.  If  an  equilateral  triangle  be  inscribed  in  a  circle,  and  a 
diameter  be  drawn  from  one  vertex,  the  triangle,  formed  by- 
joining  the  other  extremity  of  the  diameter  and  the  center  of 
the  circle  with  one  of  the  other  vertices  of  the  inscribed  tri- 
angle, will  also  be  equilateral. 

415.  The  bisectors  of  the  angles  formed  by  extending  the 
sides  of  an  inscribed  quadrilateral  are  perpendicular  to  each 
other. 

416.  If  an  equilateral  triangle  be  inscribed  in  a  circle,  and 
any  point  in  the  circumference  be  selected  at  random,  one  of 
the  lines  which  join  this  point  to  the  three  vertices  will  equal 
the  sum  of  the  other  two. 

417.  If  an  inscribed  isosceles  triangle  have  its  base  angles 
each  double  the  vertical  angle,  and  its  vertices  be  the  points  of 
contact  of  three  tangents,  these  tangents  will  form  an  isosceles 
triangle  each  of  whose  base  angles  is  one  third  its  vertical  angle. 

418.  If  through  any  point  selected  at  random  in  a  circle  a 
diameter  and  other  chords  be  drawn,  the  least  chord  will  be 
the  one  that  is  perpendicular  to  the  diameter. 

419.  If  the  sides  of  any  quadrilateral  be  the  diameters  of 
circles,  the  common  chord  of  any  adjacent  two  is  parallel  to 
the  common  chord  of  the  other  two. 

420.  If  equilateral  triangles  be  described  on  the  sides  of  any 
triangle,  the  circumferences  of  their  circumscribed  circles  are 
concurrent. 

421.  If  equilateral  triangles  be  described  on  the  sides  of  any 
triangle,  the  lines  joining  the  centers  of  these  triangles  will 
form  an  equilateral  triangle. 

Sug.     Draw  the  circumscribed  circles. 


70 


Plane  Geometry. 


PROPORTIONAL    LINES. 

422.  If  a  series  of  parallel  transversals,  intersecting  any  two  straight 
lines,  intercept  equal  distances  on  one  of  these  lines,  they  will  also  inter- 
cept equal  distances  on  the  other. 

Sug.  Make  such,  construction  as  will  enable  you  to  use 
theorems  203  and  219. 

423.  If  a  line  he  drawn  parallel  to  one  side  of  a  triangle,  the  four 
parts  into  which  it  divides  the  other  sides  will  form  a  proportion. 

B 


Fig.  I.  Fig.  II. 

Post.  Let  ABC  be  a  triangle  with  line  DF  drawn  parallel 
to  AC. 

To  Prove.    AD,  DB,  CF,  and  BF  are  proportionals. 

Case  I.    When  AD  and  DB,  Fig.  I,  are  commensurable. 

Dem.  Since  AD  and  DB  are  commensurable,  they  must  have 
a  common  measure,  as  AH. 

Then  AH  will  be  contained  in  AD  m  times  and  in  DB  n  times. 
.-.  AD:DB::m:  n.     (?) 

Now,  at  the  several  points  of  division  on  AB  suppose  lines 
are  drawn  to  CB  parallel  to  DF. 

What  will  be  the  relation  of  these  lines  to  one  another  ?  Why  ? 

These  lines  will  divide  CB  into  equal  parts.     Why  ? 

CF  will  be  divided  into  m  equal  parts  and  FB  into  n  equal 

P^^*^-  .-.   CF:FB:'.m:n.  (?) 

.♦.  AD.DB::  CFiFB.    (?) 


Proportional  Lines.  71 

Case  II.    When  AD,  and  DB,  Fig.  II,  are  incommensurable. 

Dem.  Suppose  BD  to  be  divided  into  any  number  of  equal 
parts  and  one  of  these  taken  as  a  unit  to  measure  AD.  This 
will  be  contained  a  certain  number  of  times  with  a  remainder 
AK  less  than  the  unit. 

Now  let  KH  be  drawn  parallel  to  DF. 

.-.  KD:DB::FH:BF    (By  Case  I). 

^^  DB      BF' 

Suppose  now  the  unit  of  measure  to  be  indefinitely  diminished. 

.-.  KA  would  become  a  decreasing  variable  and  DK  would 
become  an  increasing  variable  approaching  AD  as  its  limit. 

Likewise  FH  would  also  become  an  increasing  variable  with 
FC  as  its  limit. 

T^T)  TPTT 

Consequently and ,    since   their  denominators   are 

DB  BF 

constants,    become    increasing    variables    approaching    their 

AD  FC 

respective  limits  and 

^  DB  BF 

But  M^m, 

DB     BF 

.    4D^FC       .^. 
'-'    DB     BF'     ^'^ 

.-.  AD:DB::FC:BF.     (Proportion  form.)  q.e.d. 

424.  If  a  line  divide  two  sides 
of  a  triangle  into  proportional  parts, 
this  line  is  parallel  to  the  third 
side. 

Post.  Let  ACH  be  a  triangle, 
with  line  BD  drawn  so  that 

AB:BO::AD:DH. 
To  Prove.  BD  is  parallel  to  CH. 


72 


Plane  Geometry. 


Cons.     Suppose  BK  to  be  drawn  through  B  parallel  to  CH. 

Vem.                 AB:BC::AB:DH.  Why? 

AB:BC::AK:KR.  Why? 

.-.  AD  :  DH: :  AK:  KH.  Why  ? 

.-.  AD-\-DH:DH:'.  AK  +  KH:  KH.  Why  ? 

AH'.DH'.:AH:KH  Why? 

.'.DH=KH.  Why? 

.-.  points  D  and  /fmust  coincide     Why  ? 

What,  then,  must  be  the  relative  position  of  the  lines  DK 
Silid  BD?    Why? 

But  BKwsiS  drawn  parallel  to  CH.  Consequently  BD,  which 
coincides  with  BK,  must  be  parallel  to  CH  q.e.d. 

425.  If  a  hne  be  drawn  parallel  to  one  side  of  a  triangle,  the  other 
two  sides  and  either  pair  of  corresponding  parts  are  proportionals. 

Sug.     Consult  423  and  360. 

426.  If  a  line  meet  two  sides  of  a  triangle  so  that  those  two  sides 
and  either  pair  of  corresponding  parts  are  proportionals,  this  Hne  is 
parallel  to  the  third  side. 

Sug.     Proceed  as  in  424. 

427.  If  two  lines  be  drawn  parallel  to  one  side  of  a  triangle 
intersecting  the  other  two  sides,  the  parts  thus  intercepted  are 

proportionals. 

428.  If  a  line  be  drawn  parallel  to 
one  side  of  a  triangle,  this  side,  its 
parallel,  and  either  of  the  other  two 
sides,  together  with  the  part  of  the 
latter  joining  the  third  side,  are  propor- 
tionals. 

To  Prove.  BN.FR.:  AN.AR. 

Cons.        Draw  i^TF  II  to  AB. 


Proportional  Lines.  73 

429.  If  a  line  be  drawn  parallel  to  the  base  of  a  triangle, 
and  another  from  vertex  to  base,  the  parts  of  both  the  base 
and  its  parallel  are  proportionals. 

430.  The  bisector  of  an  angle  of  a  triangle  divides  the  oppo- 
site side  into  segments  proportional  to  the  other  two  sides. 

Sug.  From  the  vertex  of  the  bisected  angle  extend  one  of 
the  sides,  and  from  one  of  the  other  vertices  draw  a  line  parallel 
to  the  bisector  meeting  the  side  extended. 

431.  If  the  bisector  of  an  exterior  angle  of  a  triangle  meet 
one  of  the  sides  extended,  it  will  divide  this  side  externally  into 
segments  proportional  to  the  other  two  sides  of  the  triangle. 

Sug.  From  extremity  of  the  side  extended,  draw  a  parallel 
to  the  bisector. 

432.  If  the  median  of  a  triangle  be  drawn  and  the  angles 
which  the  median  makes  with  the  base  be  bisected,  the  line 
joining  the  extremities  of  the  bisectors  will  be  parallel  to  the 
base. 


Sug.     Consult  430  and  426. 


D 


74  Plane  Geometry. 


SIMILAR  FIGURES. 

433.  Similarity  in  general  means  likeness  of  form;  i.e.  two 
figures  are  said  to  be  similar  when  they  have  the  same  shape, 
although  they  may  differ  in  size. 

Similar  geometrical  figures  are  those  whose  homologous  angles 
are  equal,  and  whose  homologous  sides  form  a  proportion. 

434.  Homologous  sides  of  similar  figures  are  those  which 
join  the  vertices  of  equal  angles.  In  similar  triangles  the 
homologous  sides  are  those  that  are  opposite  the  equal  angles. 
The  pupil  should  be  careful  to  observe  at  the  outset  that  simi- 
larity involves  two  things,  viz.  equality  of  angles  and  propor- 
tionality of  sides. 

435.  Two  geometrical  figures  are  said  to  be  mutually  equi- 
angular when  the  angles  of  one  are  equal  respectively  to  the 
corresponding  angles  of  the  other,  and  mutually  equilateral 
when  each  side  of  one  has  an  equal  side  in  the  other. 

436.  Similar  arcs^  sectors,  and  segments,  are  those  that  corre- 
spond to  equal  central  angles. 

It  is  obvious  that  if  two  geometrical  figures  are  similar  to 
the  same  or  equal  figures,  these  two  are  similar  to  each  other. 

THEOREMS. 

437.  If  a  line  be  drawn  parallel  to  one  side  of  a  triangle,  the  tri- 
angle thus  formed  is  similar  to  the  given  triangle. 

Sug.     Consult  434,  425,  and  428. 

438.  If  two  triangles  are  mutually  equiangular,  they  are  similar. 
Sug.     Consult  437. 

439.  'If  two  angles  of  one  triangle  are  equal  respectively  to  two 
angles  of  another,  the  two  triangles  are  similar. 

440.  If  two  right  triangles  have  an  acute  angle  in  each  equal,  the 
two  triangles  are  similar. 


Similar  Figures. 


TS 


441.  If  two  triangles  have  an  angle  in  each  equal,  and  the  sides 
including  those  angles  form  a  proportion,  the  two  triangles  are  similar. 

S>ug.     Consult  437. 

442.  If  two  triangles  have  their  sides  respectively  parallel,  they 
are  similar. 

Sug.  Produce  the  non-parallel  sides  until  they  intersect, 
then  consult  114. 

443.  If  two  triangles  have  their  sides  respectively  perpendicular 
to  each  other,  they  are  similar. 

Bug.  Extend  the  ±  sides  -^ 
until  they  meet,  then  con- 
sider the  angles  of  the 
quadrilateral  NFPB  and 
the  angles  at  B.  Then 
compare  the  A  A  and  0  by 
means  of  the  A  ARK  and 
ONK. 

444.  If  two  isosceles  tri-  ^ 

angles  have  equal  vertical  angles,  they  are  similar. 

445.  If  two  triangles  be  similar,  their  homologous  altitudes  are  in 
the  same  ratio  as  either  pair  of  homologous  sides  which  include  the 
vertical  angle. 

446.  If  two  triangles  be  similar,  their  homologous  altitudes  are  in 
the  same  ratio  as  their  homologous  bases. 

447.  If  the  homologous  sides  of  two  triangles  form  a  proportion, 
the  triangles  are  similar. 

^-;r \B 


76  Plane  Geometry. 

Post.  Let  ABC  and  DHK  be  two  triangles,  tlieir  homolo- 
gous sides  forming  the  continued  proportion 

AB  :  DH:  :  ACiDK: :  BC :  HK. 

To  Prove.  That  the  two  triangles  are  similar ;  i.e.  that  they 
are  mutually  equiangular. 

Cons.  Make  JVC  equal  to  DK,  PC  equal  to  HK,  and  join 
JVP. 

Dem.  In  the  given  proportion  substitute  for  DK  and  JSK 
their  equals  JVC  and  PC     Then 

AC'.NC'.'.BC'.PC. 

What  is  true,  then,  of  the  two  triangles  ABC  and  NPC? 
See  Theorem  441. 

.-.  AB :  iVP: :  BC:  PC.     Why  ? 
But  by  Hyp.,  AB :  D^:  :  BC:  KH. 

What  is  true  of  the  first  terms  of  these  two  proportions? 
Of  the  third  terms  ?  Of  the  last  terms  ?  What  must  be  true, 
then,  of  the  second  terms  ? 

The  pupil  should  finish  the  demonstration  without  diffi- 
culty. 

448.  If,  in  a  right  triangle,  a  perpendicular  be  drawn  to  the 
hypothenuse  from  the  vertex  of  the  right  angle,  — 

i.   The  two  triangles  thus  formed  are  similiar  to  the  original 
triangle  and  to  each  other. 

ii.   The  perpendicular  and  the  two  segments  of  the  hypothenuse  are 
proportionals. 

iii.  Either  leg,  the  hypothenuse,  and  that  segment  of  the  latter 
which  joins  the  former,  are  proportionals. 

iv.  The  two  segments  of  the  hypothenuse  and  the  squares  of  the  two 
legs  are  proportionals. 

Sug.  Use  the  two  proportions  of  iii.  by  Theorem  356,  and 
divide  one  by  the  other. 


Similar  Figures.  "jj 

V.  The  square  of  the  hypothenuse  bears  the  same  ratio  to  the 
square  on  either  leg  as  the  hypothenuse  bears  to  that  segment  of  the 
latter  joining  that  leg. 

&ug.     Consult  Part  iii.  and  378. 

vi.  The  two  legs,  the  hypothenuse,  and  the  perpendicular  are 
proportionals. 

Bug,  Select  the  two  similar  triangles  and  pick  out  the 
homologous  sides. 

449.  If  from  any  point,  in  the  circumference  of  a  circle,  a  per- 
pendicular be  drawn  to  a  diameter,  this  perpendicular  will  be  a  mean 
proportional  between  the  two  segments  of  the  diameter. 

k^ug.     Consult  394. 

450.  If  two  chords  of  a  circle  intersect  each  other,  the  four  parts 
are  proportionals. 

Bug.    Draw  other  chords  and  compare  the  triangles. 

451.  If  two  secants  be  drawn  from  the  same  point  without  the 
circle,  the  entire  secants  and  their  external  segments  are  proportionals. 

Bug.     Draw  such  chords  as  will  make  similar  triangles. 

452.  If,  from  the  same  point,  a  tangent  and  secant  to  a  circle  be 
drawn,  the  tangent,  the  secant,  and  the  external  segment  of  the  latter, 
are  proportionals. 

453.  If  two  circles  intersect  each  other,  the  two  tangents 
from  any  point  in  their  common  secant  are  equal. 

454.  If  two  circles  intersect  each  other,  their  common  secant 
bisects  their  common  tangents. 


78  Plane  Geometry. 


POLYGONS. 

455.  A  Polygon  is  a  portion  of  a  plane  bounded  by  straight 
lines. 

What  is  the  least  number  of  sides  that  a  polygon  can  have  ? 
The  greatest  number  ? 

456.  The  word  ^polygon  is  from  two  Greek  words  meaning 
many  angles. 

What  is  a  polygon  of  three  sides  usually  called  ?  It  is  also 
called  a  trigon. 

A  polygon  of  four  sides  is  usually  termed  what  ?  It  is  also 
called  a  tetragon. 

A  polygon  of  five  sides  is  called  a  pentagon^  one  of  six  sides 
a  hexagon,  one  of  seven  sides  a  heptagon,  one  of  eight  sides  an 
octagon,  one  of  nine  sides  a  nonagon  or  enneagon,  one  of  ten 
sides  a  decagon,  one  of  eleven  sides  an  undecagon,  one  of  twelve 
sides  a  dodecagon,  and  one  of  fifteen  sides  a  peritedecagon. 

457.  A  Convex  Polygon  is  one  each  of  whose  angles  is  less 
than  180°,  and 

A  Concave  Polygon  is  one  that  has  one  or  more  reflex 
angles. 

Whenever  polygons  are  mentioned  in  this  work,  convex 
polygons  are  meant  unless  otherwise  specified. 

The  Diagonal  of  a  polygon  is  a  line  joining  any  two  vertices 
not  consecutive. 


458.  A  Regular  Polygon  is  one  that  is  both  equilateral  and 
equiangular. 

459.  A  polygon  is  inscribed  in  a  circle  when  the  vertices  of 
all  its  angles  are  on  the  circumference ;  i.e.  when  its  sides  are 
chords  of  that  circle.  The  circle  is  said  then  to  be  circum- 
scribed about  the  polygon,  and  the  radius  of  this  circle  is  also 
called  the  radius  of  the  polygon. 


Polygons. 


79 


460.  A  polygon  is  circumscribed  about  the  circle  when  its 
sides  are  all  tangents  to  the  circle.  The  circle  is  then  said  to 
be  inscribed  in  the  polygon,  and  the  radius  of  this  circle  is 
called  the  apothem  of  the  polygon. 

461.  The  Perimeter  of  a  polygon  is  the  sum  of  its  sides. 
How  many  angles  has  each  of  the  above-named  polygons? 
How  does  the  number  of  sides  compare,  in  each  case,  with  the 
number  of  angles  ?  How  many  angles,  then,  has  a  polygon  of 
n  sides  ?  How  many  diagonals  can  be  drawn  from  a  single 
vertex  in  each  of  the  above-named  polygons?  Compare  the 
number  of  diagonals  in  each  case  with  the  number  of  sides. 
If  the  polygon  has  n  sides,  how  many  diagonals  can  be  drawn 
from  a  single  vertex?  How  many  triangles  are  formed  by 
drawing  diagonals  from  a  single  vertex  in  each  of  the  above- 
named  polygons?  How  does  the  number  of  triangles  com- 
pare with  the  number  of  sides  ?  If  the  polygon  has  n  sides, 
how  many  triangles  would  be  formed  by  diagonals  similarly 
drawn  ? 

Fill  out  the  following  table : 


Diagram 

Jfo.  of  sides  of 
the  poIyRon 

No.  of  diagonals 
drawn  from  one  vertex 

No.  of,^    formed 

Value  of  Z  8  In  rt  Z,  a 

3 

4 

6 

6 

7 

8 

j 

; 

n 

462.  The  sum  of  the  interior  angles  of  a  polygon  of  n  sides  is  equal 
to  twice  as  many  right  angles  as  the  polygon  has  sides,  less  four  right 
angles;  i.e.  2  22  —4  right  angles. 

Sug.  Let  n  equal  the  number  of  sides  and  consult  the  above 
table. 


8o 


Plane  Geometry. 


Cor.  If  a  polygon  of  n  sides  is  equiangular,  each  angle  is 
equal  to  ^^Lzii  rt.  A, 


463.  Tlie  sum  of  tlie  exterior  angles  of  any  polygon  formed  by  ex- 
tending each  of  its  sides  in  succession 
similarly,  is  equal  to  four  right  angles. 

Sug. 

Int.  A  +  Ext.  A  =  2n. 
Int.  A  =2  71-4:. 


.'.  Ext.  A  =  4. 


Q.E.D. 


464.  The  bisectors  of  the  angles  of  a  regular  polygon  are  concurrent. 

Post.  Let  ABW,  etc.,  be  a  regu- 
lar polygon,  and  let  AO  and  BO 
bisect  A  A  and  B,  and  let  OWhe 
drawn. 

To  Prove.     OTT  bisects  Z  W. 
Cons.     Draw  Js   OJSFj   OF,   and 
OQ. 
Dem.   Compare  the  A. 

465.  The  perpendicular  bisectors  of 
the  sides  of  a  regular  polygon  are 
concurrent  in  the  same  point  as  the  angle  bisectors. 

Sug.   Draw  angle  bisectors  and  Js  to  the  sides,  then  use 
above  diagram. 

466.  If  an  equiangular  polygon  be  circumscribed  about  a  circle,  the 
bisectors  of  its  angles  are  concurrent  at  the  center  of  the  circle. 

Sug.   Consult  328. 

467.  If  an  equilateral  polygon  be  inscribed  in  a  circle,  it  is  regular. 
Sug.   Consult  392. 


Polygons. 


8i 


468.  If  an  equiangular  polygon  be  circumscribed  about  a  circle,  it 
is  regular. 

Sug.   Consult  466. 

469.  If  a  polygon  be  regular,  a  circle  can  be  circumscribed  about  it. 

470.  If  a  polygon  be  regular,  a 
circle  can  be  inscribed  in  it. 

471.  One  side  of  a  regular  hexagon 
is  equal  to  the  radius  of  the  circum- 
scribed circle. 

Sug,  Prove  ANK  an  equilateral 
triangle  by  first  comparing  the 
angles. 

472.  If  the  circumference  of  a  circle 
be  divided  into  any  number  of  equal 
arcs,  the  chords  joining  the  successive 
points  of  division  will  form  a  regular  polygon. 

473.  If  a  regular  polygon  be  circumscribed  about  a  circle,  the  points 
of  contact  will  divide  the  circumference  into  equal  arcs. 

474.  If  the  circumference  of  a  circle  be  divided  into  any  number  of 
equal  arcs,  the  tangents  at  the  points  of  division  will  form  a  regular 
polygon. 

475.  Tangents  to  a  circumference  at  the  vertices  of  a  regular 
inscribed  polygon  form  a  regular  circumscribed  polygon  of  the  same 
number  of  sides  as  the  inscribed  polygon. 

476.  Two  regular  polygons  of  the  same  number  of  sides  are  similar. 

X b: 

IV, T 


82  Plane  Geometry. 

Post.  Let  P  and  P  be  two  regular  polygons,  each  having  n 
sides. 

To  Prove  that  they  are  similar;  i.e.  that  their  homologous 
sides  form  a  proportion,  and  that  they  are  mutually  equi- 
angular. 

Sug.  Compare  AB  and  BC,  etc.,  and  NQ  and  QR  etc.,  and 
thus  obtain  a  proportion  j  then  compare  the  A  A  and  N,  B  and 
Q,  etc. 

477.  If  two  polygons  be  similar,  the  diagonals  drawn  from  homolo- 
gous vertices  will  divide  them  into  the  same  number  of  trangles,  similar 
two  and  two,  and  similarly  placed. 

Sug.   Consult  441. 

478.  If  the  diagonals  drawn  from  homologous  vertices  of  two  poly- 
gons divide  them  into  the  same  number  of  triangles,  similar  two  and 
two,  and  similarly  placed,  the  two  polygons  are  similar. 

Sug.   Consult  362. 

479.  The  diagonals  drawn  from  homologous  vertices  of  two  similar 
polygons  are  proportionals. 

480.  The  perimeters  of  two  similar  polygons  are  in  the  same  ratio 
as  any  two  homologous  sides  and  as  any  two  homologous  diagonals. 

Sug.  Consult  367. 

481.  The  perimeters  of  two  similar  polygons  are  in  the  same  ratio  as 
the  bisectors  of  any  two  homologous  angles. 

482.  The  perimeters  of  two  regular  polygons  of  the  same  number  of 
sides  are  in  the  same  ratio  as  the  radii  of  their  circumscribed  circles. 

Sug.   Consult  444  and  480. 

483.  The  perimeters  of  two  regular  polygons  of  the  same  number 
of  sides  are  in  the  same  ratio  as  their  apothems. 

PROBLEMS  OF   COMPUTATION. 

484.  Compute  the  value  in  right  angles  of  the  sum  of  the 
interior  angles  of  a  pentagon,  hexagon,  octagon,  decagon,  dodeca- 
gon, pentedecagon,  and  a  polygon  oiffiy-tivo  sides. 


Polygons.  83 

485.  Compute  the  value,  in  terms  of  a  right  angle  as  the 
unit,  of  one  of  the  angles  of  an  equiangular  peritagon,  octagon, 
dodecagon,  a  polygon  of  twenty  sides,  one  hundred  sides. 

486.  Compute  values  of  the  above  angles  in  degrees. 

487.  How  many  sides  has  an  equiangular  polygon,  the  sum 
of  four  angles  of  which  is  equal  to  seven  right  angles  ? 

488.  How  many  sides  has  an  equiangular  polygon,  the  sum 
of  three  angles  of  which  is  equal  to  five  right  angles  ? 

489.  How  many  sides  has  an  equiangular  polygon,  the  sum 
of  nine  angles  of  which  is  equal  to  sixteen  right  angles  ? 

490.  How  many  sides  has  the  polygon,  the  sum  of  whose 
interior  angles  is  equal  to  the  sum  of  its  exterior  angles  ? 

491.  How  many  sides  has  the  polygon,  the  sum  of  whose 
interior  angles  is  double  that  of  its  exterior  angles  ? 

492.  How  many  sides  has  the  polygon,  the  sum  of  whose 
exterior  angles  is  double  that  of  its  interior  angles  ? 

493.  How  many  sides  has  the  polygon,  the  sum  of  whose 
interior  angles  is  equal  to  nine  times  the  sum  of  its  exterior 
angles  ? 

494.  How  many  sides  has  the  equiangular  polygon,  when 
the  sum  of  nine  of  its  interior  angles  is  four  times  the  sum  of 
its  exterior  angles  ? 

495.  How  many  sides  has  the  equiangular  polygon,  when 
the  sum  of  five  of  its  interior  angles  is  equal  to  two  and  one 
fourth  times  the  sum  of  its  exterior  angles  ? 

496.  How  many  sides  has  the  equiangular  polygon,  when 
the  sum  of  four  of  it's  interior  angles  equals  seven  and  one  half 
right  angles  ? 

497.  How  many  sides  has  the  equiangular  polygon,  when 
the  sum  of  six  of  its  interior  angles  equals  three  and  three 
fourths  times  the  sum  of  its  exterior  angles  ? 


84 


Plane  Geometry. 


PROJECTION  AND   AREAS. 

498.  The  Projection  of  one  line  upon  another  is  that  portion 
of  the  latter  included  between  two  perpendiculars  to  the  latter 
drawn  from  the  extremities  of  the  former. 


Fig.  I. 


Fig.  II. 


Fig.  III. 


For  example,  UK  is  the  projection  of  the  line  CD  upon  AB, 
and  BS  is  the  projection  of  BQ  on  iVP;  it  is  also  the  pro- 
jection of  BQ  on  RF.  Mention  all  the  cases  of  projection  in 
Fig.  Ill,  the  dotted  lines  being  perpendicular  to  the  respective 
sides. 

499.  The  Area  of  a  surface  is  its  numerical  measure ;  i.e.  the 
numerical  expression  for  the  number  of  times  it  contains  an- 
other surface  arbitrarily  assumed  as  a  unit  of  measure.  For 
example,  the  area  of  the  floor  of  a  room  is  the  number  of  times 
it  contains  some  one  of  the  common  units  of  surface,  square 
foot,  square  yard,  etc. 

This  unit  of  surface  is  called  the  superficial  unit.  The  most 
convenient  superficial  unit  is  the  square  of  the  linear  unit. 
The  square  of  the  linear  unit  (or  line)  is  the  surface  of  the 
square  constructed  with  that  linear  unit  for  its  sides. 

Surfaces  that  are  equal  in  area  are  said  to  be  equivalent. 

500.  By  the  product  of  two  lines  is  meant  the  product  of  the 
numbers  which  represent  them  when  both  are  measured  by 
the  same  linear  unit.  It  is  also  sometimes  expressed  as  the 
rectangle  of  two  lines. 


Projection  and  Areas. 


501.   If  two  rectangles  liave  equal  altitudes,  tlieir 
the  same  ratio  as  their  bases. 

Case  I.   When  the  bases  are  commensurable. 


85 

will  be  in 


Q 


B 

2J- 


B     H 

Post.  Let  ABCD  and  HKNP  be  two  rectangles,  having  their 
altitudes  AD  and  HP  equal,  and  their  bases  AB  and  HK  com- 
mensurable.    Designate  their  areas  by  Q  and  R  respectively. 

To  Prove.     Q:B::AB:  HK. 

Dem.  Since  AB  and  HK  are  commensurable,  they  must 
have  a  common  measure.  (^31) 

Suppose  this  common  measure  to  be  contained  in  AB  n  times, 
and  in  HK  m  times. 

At  the  several  points  of  division  construct  Js  to  AB  and  HK 
Then  the  rectangle  ABCD  will  be  divided  into  n  equal  rectan- 
gles, and  the  rectangle  HKNP  will  be  divided  into  m  equal  rec- 
tangles.   These  smaller  rectangles  are  also  equal  to  one  another. 

Hence  AB:KH::n:m,  (?) 

and  Q:B::n:m.  (?) 

.-.  Q:E::AB:HK.     (?) 

Case  II.   When  the  bases  are  incommensurable. 


W  N 


Dem.  Suppose  AB  to  be  divided  into  any  number  of  equal 
parts  as  n,  and  that  one  of  these  parts  be  applied  to  HK  and 
found  to  be  contained  m  times  with  a  remainder  less  than  that 


86 


Plane  Geometry. 


part.  From  the  points  of  division  construct  lines  as  in  Case  I. 
Then  the  rectangle  ABGD  will  be  divided  into  n  equal  rectan- 
gles, and  rectangle  NKHP  into  m  equal  rectangles,  with  a 
remaining  rectangle  less  than  one  of  the  m  rectangles. 

Then,   Rectangle  Q :  rectangle  POi.AB:  HO.     (By  Case  I.) 
If,  now,  the  unit  of   division  be  indefinitely  diminished, 
base  HO  and  Eect.  PO  become  increasing  variables  approach- 
ing their  respective  limits  KH  and  B. 

.'.  — — - — ^becomes  a  decreasing  variable  approaching  its 

limit  ^^^  •  ^,  and  — -  becomes  a  decreasing  variable  approach- 
llect.  li  HO 

AB 

ing  its  limit  — — • 

xzit 


But 


or 


Rect.    Q 
Eect.  PO 


AB 
HO 


Rect.  Q^AB 
Rect.  E     HK' 


(?) 


Rect.  Q :  Rect.  R : :  ^5 :  HK 


Q.E.D. 


502.  If  two  rectangles  have  equal  bases,  their  areas  are  in  the  same 
ratio  as  their  altitudes. 

Sug.     Consider  their  altitudes  as  bases,  and  bases  as  alti- 
tudes, and  proceed  as  in  501. 

503.  The  areas  of  any  two  rectangles  have  the  same  ratio  as  the 
products  of  their  respective  bases  and  altitudes. 


W^ 


Fig.  I. 


K 


N 


M       O 


O 


-J, 


Fig.  II. 


Projection  and  Areas.  87 

Post  Let  RVCD  and  WKNP  be  any  two  rectangles  with 
bases  DC  and  PN.  Designate  their  areas  by  A  and  a  re- 
spectively, their  altitudes  by  H  and  h,  and  their  bases  by 
B  and  h. 

To  Prove.     AiaiiB-Hib-  h. 

Cons.  Draw  FG  =  DC  and  prolong  it,  making  GJ=  KN. 
Through  G  draw  MQ  ±  to  FJ,  making  MG  =  RD,  and  GQ  = 
WK.     Complete  the  three  rectangles  as  shown  in  Fig.  II. 

Sug.     Consult  501  and  373. 

504.  Hence,  the  above  demonstration  shows  that  if  the 
bases  and  altitudes  of  any  two  rectangles  be  measured  by  the 
same  linear  unit,  the  ratio  of  their  respective  products  (see 
500)  will  be  the  ratio  of  their  respective  surfaces ;  and  conse- 
quently either  may  be  assumed  as  the  measure  of  the  other. 

For  example,  suppose  the  linear  unit  to  be  contained  5  times 
in  ADy  18  times  in  DCy  3  times  in  HP,  and  6  times  in  PN. 

Then,  ^:  a:  :5 -18:3 -6, 

or  —  =  — • 

a     18 

This  simply  means  that,  using  a  as  the  unit,  the  area  A  is 
5  times  as  large  as  the  area  a ;  or  that  the  area  a,  using  A  as 
the  unit  surface,  is  \  as  large  as  A. 

Practically  it  is  more  convenient  to  compare  the  areas  of 
both  rectangles  with  square  of  the  linear  unit  (499)  as  a  unit 
of  surface.  This  comparison  is  formally  shown  in  the  enuncia- 
tion and  demonstration  of  the  following  theorem. 

505.  The  area  of  a  rectangle  is  equal  to  the  product  of  its  base  and 
altitude. 

Post.     Let  ABCD  be  any  ^' ^ 

rectangle;  and  in  whatever       1 
linear  unit  the  base  and  alti-    \^  \l 
tude  be  expressed,  let  r  be  a 
square  whose  sides  are  the 


88 


Plane  Geometry. 


same  unit.    Designate  its  area  by  Ry  and  its  base  and  altitude 

by  h  and  h  respectively. 

To  Prove.  It  =  b-h. 


Dem. 


or 


i?:r::6./i:l.l, 

R 

r 

or,  since  r  is  the  unit  of  surface 


(503) 


1.1^ 


Q.E.D. 


506.  When  the  base  and  altitude  are  commensurable,  this  is 
rendered  evident  by  dividing  the  rectangle  into  squares,  each 
equal  to  the  superficial  unit,  as  shown  in  annexed  figure.  The 
above  demonstration,  however,  includes  the  case  when  the  base 
and  altitude  are  incommensurable,  hence, 


"{- 


507.  If  two  rectangles  have  the  same  or  equal  bases  and  the  same 
or  equal  altitudes,  their  areas  are  equal. 

508.  If  a  parallelogram  and  rectangle  have  the  same  or  equal  bases 
and  the  same  or  equal  altitudes,  they  are  equivalent. 


Post. . 

To  Prove.     . 

.  Sug.    Extend  CD,  draw  BO  and  AP  ±  to  AB^  then  com- 
pare ABOP  with  FHKN  and  ABCD. 


Projection  and  Areas.  89 

509.  The  area  of  a  parallelogram  is  equal  to  the  product  of  its  base 
and  altitude. 

510.  If  two  parallelograms  have  the  same  or  equal  bases  and  the 
same  or  equal  altitudes,  they  are  equivalent. 

511.  The  areas  of  any  two  parallelograms  are  in  the  same  ratio  as 
the  products  of  their  respective  bases  and  altitudes. 

512.  If  two  parallelograms  have  the  same  or  equal  bases,  their  areas 
are  in  the  same  ratio  as  their  altitudes. 

513.  If  two  parallelograms  have  the  same  or  equal  altitudes,  their 
areas  are  in  the  same  ratio  as  their  bases. 

514.  If  a  triangle  and  parallelogram  have  the  same  or  equal  bases 
and  the  same  or  equal  altitudes,  the  area  of  the  latter  is  double  that  of 
the  former. 

tSug.  Complete  the  parallelogram  having  the  triangle  for 
one  half. 

515.  The  area  of  a  triangle  is  equal  to  one  half  the  product  of  its 
base  and  altitude. 

516.  If  two  triangles  have  the  same  or  equal  bases  and  the  same  or 
equal  altitudes,  they  are  equivalent. 

517.  The  areas  of  any  two  triangles  are  in  the  same  ratio  as  the 
products  of  their  respective  bases  and  altitudes. 

518.  If  two  triangles  have  the  same  or  equal  bases,  their  areas  are 
in  the  same  ratio  as  their  altitudes. 

519.  If  two  triangles  have  the  same  or  equal  altitudes,  their  areas 
are  in  the  same  ratio  as  their  bases. 

520.  The  area  of  a  trapezoid  is  equal  to  the  product  of  its  altitude 
and  one  half  the  sum  of  its  parallel  sides. 

521.  The  area  of  a  rhombus  is  equal  to  one  half  the  product  of 
its  diagonals. 

522.  The  square  described  upon  the  sum  of  two  lines  is 
equivalent  to  the  sum  of  the  squares  upon  the  two  lines,  plus 
twice  the  rectangle  formed  by  the  two  lines. 


90  Plane  Geometry. 

523.  The  square  described  upon  the  difference  of  two  lines 
is  equivalent  to  the  sum  of  the  squares  upon  the  two  lines, 
minus  twice  the  rectangle  formed  by  the  two  lines. 

524.  The  rectangle  formed  by  the  sum  and  difference  of 
two  lines  is  equivalent  to  the  difference  of  the  squares  upon 
the  two  lines. 

625.  The  square  described  upon  the  hypothenuse  of  a  right  triangle 
is  equivalent  to  the  sum  of  the  squares  described  upon  the  legs. 

JSch.  This  theorem  was  first  demonstrated  by  Pythagoras, 
about  450  b.c,  and  hence  is  called  the  Pythagorean  theorem. 
It  has  been  a  favorite  one  with  mathematicians,  and  conse- 
quently about  fifty  different  demonstrations  of  it  have  been 
recorded.  Among  the  many  diagrams  used,  the  following  are 
a  few ;  and  it  is  hoped  that  the  pupil  will  endeavor  to  demon- 
strate it  for  himself,  using  either  one  of  these,  or,  preferably, 
one  of  his  own.  Fig.  I  is  a  diagram  used  by  Euclid  in  the 
demonstration  of  this  theorem,  constituting  his  famous  "  47th." 
Fig.  yil  is  the  diagram  used  by  the  late  President  Garfield, 
who  was  said  to  have  utilized  his  leisure  hours  in  Congress  in 
mathematical  investigations.  In  each  figure  ABC  is  the  given 
triangle,"  and  in  Fig.  VII  ADC  is  one  half  the  square  upon 
the  hypothenuse.  DH  is  drawn  parallel  to  BC  to  meet  BA 
extended.  The  method  is  algebraic,  and  involves  an  equa- 
tion between  the  sum  of  the  areas  of  the  three  triangles  and 
the  trapezoid  BCDH. 

In  Fig.  IX  consult  452  and  356.  Then^D  =  AB+CB  and 
AF^AB-CB. 

In  Fig.  XIII,  PQ  and  HF  are  drawn  parallel  to  AB,  and 
ON  parallel  to  AG.  Points  H  and  N,  as  also  K  and  iV,  are 
joined.  The  lines  iViffand  NK  can  then  easily  be  proved  to  be 
one  line.  Then  compare  the  areas  AF  and  CF,  also  BN  and 
AF.  Compare  also  the  areas  PC  and  PB,  AD  and  00,  then 
OC  and  PB. 


Projection  and  Areas. 


91 


Fig.  I. 


Fig.  II. 


Fia.  m 


Fig.  V. 


Fig.  Vn. 


Fig.  VI. 


92 


Plane  Geometry. 


Fig.  VIII. 


Fia.  XII. 


V --^ 

B 

C 

Fig.  XI. 


Fig.  XIII. 


526.  If  the  square  described  on  one  side  of  a  triangle  is  equivalent 
to  the  sum  of  the  squares  upon  the  other  two  sides,  then  the  triangle  is 
a  right  triangle. 

Post.    JjQt  AT^  =^  Iff  +  BG". 


Projection  and  Areas. 

B 


93 


A  C 

To  Prove.    Z  ACB  is  a  rt.  Z. 

Cons.     Draw  CF  ±  to  BC  and  equal  to  AC, 

Sug.     Consult  525,  Ax.  iii,  and  95. 

527.  The  square  described  upon  the  diagonal  of  a  square  is 
equivalent  to  twice  the  square  of  one  of  its  sides. 

528.  The  diagonal  of  a  square  is  incommensurable  with  its 
side. 

Sug.    Prove  it  equal  to  V2. 

529.  In  any  triangle  the  square  of  the  side  opposite  an  acute  angle 
is  equal  to  the  sum  of  the  squares  of  the  other  two  sides,  minus  twice 
the  product  of  one  of  those  sides  and  the  projection  of  the  other  upon 
that  side. 

If  C  be  the  acute  angle,  then  by  a  glance  at  the  following 
diagram  it  will  be  readily  seen  that  there  may  be  two  cases 
depending  upon  whether  the  projection  involves  an  extension 
of  one  side,  which  will  evidently  be  the  case  if  the  side  to  be 
projected  is  opposite  an  obtuse  angle. 

A  A 


B 

To  Prove. 
In  Case  I. 
In  Case  II. 


C 

aS  =  B&  +  AC' 
DB  =  BC-DC. 
DB  =  DC-BC. 


94 


Plane  Geometry. 


The  square  of  either  of  these  two  equations  gives  the  same 
result;  hence,  to  the  square  add  the  square  of  the  perpen- 
dicular, and  then  combine  the  terms  by  using  Theorem  525. 

JSch.  The  above  demonstration  is  algebraic  and  is  given 
preference  merely  because  of  a  less  complex  diagram.  The 
diagram  for  the  geometrical  demonstration  is  herewith  ap- 
pended, however,  with  the  requisite  auxiliary  lines  drawn. 
The  latter  are  drawn  similar  to  those  in  525. 


530.  In  an  obtuse  triangle  the  square  of  the  side  opposite  the 
obtuse  angle  is  equal  to  the  sum  of  the  squares  of  the  other  two 
sides,  plus  twice  the  product  of  one  of  those  sides  and  the  projection 
of  the  other  upon  that  side. 

Siig.  Form  an-  equation  by  placing  the  projection  of  the 
side  opposite  the  obtuse  angle  equal  to  the  sum  of  its  two 


Projection  and  Areas. 


95 


parts,  then  proceed  as  in  529,  or  the  following  diagram  may  be 
used  for  a  geometrical  demonstration. 


531.  If  any  median  of  a  triangle  be  drawn, 

i.  The  sum  of  the  squares  of  the  other  two  sides  is  equal  to  twice 
the  square  of  one  half  the  bisected  side,  plus  twice  the  square  of  the 
median;  and 

ii.  The  difference  of  the  squares  of  the  other  two  sides  is  equal  to 
twice  the  product  of  the  bisected  side  and  the  projection  of  the  median 
upon  that  side. 

Sug.     Use  529  and  530,  then  combine  the  resulting  equations. 

532.  If  two  triangles  have  an  angle  in  each  equal,  their  areas  are  in 
the  same  ratio  as  the  products  of  the  sides  which  include  the  equal  angles. 

Post.     Let  ABC  and  DHKhe  two  triangles  with  ZA=:ZH, 
Designate  their  areas  by  S  and  s,  and  the  sides  opposite  the 
several  angles  by  the  corresponding  small  letters. 


g6  Plane  Geometry. 


To  Prove.  iS :  s::c  -h:  :k  '  d. 

Cons.  Extend  the  sides  of  one  until  they  equal  the  corre- 
sponding sides  of  the  other,  as  in  the  above  diagram,  and  join  JSfK. 

Dem.     Area  DIIK :  Area  NHK  ::DH:  NH.     (519) 
Area  NHK :  Area  NHP  : :  HK :  HP.     Why  ? 

Whence 

AiGSi  DHK :  Area.  NHP  I'.DHxHKiHJSTx  HP.    Why? 
or  s  :  S  : :  k  •  d  :  c  -  b.  q.e.d. 

533.  The  areas  of  two  similar  triangles  have  the  same  ratio  as  the 
squares  of  any  two  homologous  sides. 

JSug.     Consult  532,  372,  and  362. 

534.  The  areas  of  two  similar  triangles  are  in  the  same  ratio  as  the 
squares  of  their  homologous  altitudes. 

635.   The  areas  of  two  similar  polygons  are  in  the  same  ratio  as  the 
squares  of  any  two  homologous  sides. 
JSug.     Consult  477,  533,  and  367. 

536.  The  areas  of  two  similar  polygons  are  in  the  same  ratio  as  the 
squares  of  any  two  homologous  diagonals, 

537.  Any  two  homologous  sides  or  diagonals  of  two  similar 
polygons  are  in  the  same  ratio  as  the  square  roots  of  their  areas. 

538.  The  areas  of  two  regular  polygons  of  the  same  number 
of  sides  are  in  the  same  ratio  as  the  squares  of  any  two  homolo- 
gous sides. 

Sug.     Consult  476. 


Theorems.  97 

639.  The  areas  of  two  regular  polygons  of  the  same  number  of  sides 
are  in  the  same  ratio  as  the  squares  of  the  radii  of  their  circumscribed 
circles. 

540.  The  areas  of  two  regular  polygons  of  the  same  number  of  sides 
are  in  the  same  ratio  as  the  squares  of  their  apothems. 

541.  The  area  of  a  regular  polygon  is  equal  to  one  half  the  product 
of  its  perimeter  and  apothem. 

Sug.   Draw  radii  and  apothems,  then  consult  515. 

ADVANCE    THEOREMS. 

542.  In  any  triangle  the  product  of  any  two  sides  is  equal 
to  the  diameter  of  the  circumscribed  circle  multiplied  by  the 
perpendicular  drawn  to  the  third  side  from  the  vertex  of 
the  angle  opposite. 

Sug.  Construct  the  diameter  from  the  same  vertex  as  the 
perpendicular,  and  join  its  extremity  with  one  of  the  other 
vertices,  making  a  right  triangle  similar  to  the  one  formed  by 
the  perpendicular,  whence  the  necessary  proportion. 

543.  In  any  triangle  the  product  of  any  two  sides  is  equal 
to  the  product  of  the  segments  of  the  third  side,  formed  by 
the  bisector  of  the  opposite  angle,  plus  the  square  of  the 
bisector. 

Sug.  Circumscribe  a  circle,  and  extend  the  bisector  to  the 
circumference,  and  connect  its  extremity  with  one  of  the  other 
vertices  of  the  triangle.  If  the  vertices  of  the  triangle  be 
lettered  A,  B,  and  O,  and  the  bisector  be  drawn  from  A,  and  D 
the  point  where  the  bisector  crosses  the  side  BC,  and  E  the 
extremity  of  the  bisector  extended,  then  the  triangles  BAD  and 
ACE  can  be  proved  similar. 

544.  If  one  leg  of  a  right  triangle  is  double  the  other,  the 
perpendicular  from  the  vertex  of  the  right  angle  to  the 
hypothenuse  divides  it  into  segments  which  are  to  each 
other  as  1  to  4. 


98  Plane  Geometry. 

546.  A  line  parallel  to  the  bases  of  a  trapezoid,  passing 
through,  the  intersection  of  the  diagonals,  and  terminating  in 
the  non-parallel  sides,  is  bisected  by  the  diagonals. 

546.  In  any  triangle  the  product  of  any  two  sides  is  equal 
to  the  product  of  the  segments  of  the  third  side,  formed  by 
the  bisector  of  the  exterior  angle  at  the  opposite  vertex, 
minus  the  square  of  the  bisector. 

iSug.     Circumscribe  a  circle  as  in  543. 

547.  The  perpendicular  from  the  intersection  of  the  medians 
of  a  triangle,  upon  any  straight  line  in  the  plane  of  the 
triangle,  is  one  third  the  sum  of  the  perpendiculars  from  the 
vertices  of  the  triangle  upon  the  same  line. 

548.  If  two  circles  are  tangent  to  each  other,  their  common 
tangent  and  their  diameters  form  a  proportion. 

549.  If  two  circles  are  tangent  internally,  all  chords  of  the 
greater  circle  drawn  from  the  point  of  contact  are  divided 
proportionally  by  the  circumference  of  the  smaller. 

550.  In  any  quadrilateral  inscribed  in  a  circle,  the  product 
of  the  diagonals  is  equal  to  the  sum  of  the  products  of  the 
opposite  sides. 

JSug.  From  one  vertex  draw  a  line  to  the  opposite  diagonal, 
making  the  angle  formed  by  it  and  one  side  equal  to  the 
angle  formed  by  the  other  diagonal  and  side  which  meets 
the  former. 

551.  If  three  circles  whose  centers  are  not  in  the  same 
straight  line  intersect  one  another,  the  common  chords  are 
concurrent. 

552.  If  two  chords  be  perpendicular  to  each  other,  the 
sum  of  the  squares  of  the  four  segments  is  equal  to  the  square 
of  the  diameter. 

553.  The  sum  of  the  squares  of  the  diagonals  of  a  quadri- 
lateral is  equal  to  twice  the  sum  of  the  squares  of  the  lines 
joining  the  middle  points  of  the  opposite  sides. 


Theorems.  99 

554.  If  two  triangles  have  two  sides  of  one  equal  respec- 
tively to  two  sides  of  the  other,  and  their  included  angles 
supplementary,  the  triangles  are  equivalent. 

555.  If  a  straight  line  be  drawn  through  the  center  of  a 
parallelogram,  the  two  parts  are  equivalent. 

556.  If  through  the  middle  point  of  the  median  of  a 
trapezoid  a  line  be  drawn,  cutting  the  bases,  the  two  parts 
are  equivalent. 

557.  In  every  trapezoid  the  triangle  which  has  for  its  base 
one  leg,  and  for  its  vertex  the  middle  point  of  the  other  leg, 
is  equivalent  to  one  half  the  trapezoid. 

558.  If  any  point  within  a  parallelogram,  selected  at  ran- 
dom, be  joined  to  the  four  vertices,  the  sum  of  the  areas  of 
either  pair  of  opposite  triangles  is  equivalent  to  one  half  the 
parallelogram. 

559.  The  area  of  a  trapezoid  is  equal  to  the  product  of 
one  of  its  legs  and  the*  distance  of  this  leg  from  the  middle 
point  of  the  other. 

560.  The  triangle  whose  vertices  are  the  middle  points  of 
the  sides  of  a  given  triangle  is  equivalent  to  one  fourth  the 
latter. 

561.  In  any  quadrilateral  the  sum  of  the  squares  of  the 
four  sides  is  equal  to  the  sum  of  the  squares  of  the  diagonals, 
plus  four  times  the  square  of  the  line  joining  the  middle  points 
of  the  diagonals. 

Sug.    Draw  such  lines  as  will  utilize  531. 
If  the  quadrilateral  were  a  parallelogram,  how  would  that 
modify  the  above  theorem  ? 

562.  If  two  parallelograms  have  two  contiguous  sides  respec- 
tively equal,  and  their  included  angles  supplementary,  the 
parallelograms  are  equivalent. 


lOO  Plane  Geometry. 

663.  The  lines  joining  the  middle  point  of  either  diagonal  of 
a  quadrilateral  to  the  opposite  vertices,  divide  the  quadrilateral 
into  two  equivalent  parts. 

564.  The  line  which  joins  the  middle  points  of  the  bases 
of  a  trapezoid  divides  the  trapezoid  into  two  equivalent  parts. 

PROBLEMS  OF  COMPUTATION. 

565.  The  chord  of  one  half  a  certain  arc  is  9  inches,  and 
the  distance  from  the  middle  point  of  this  arc  to  the  middle  of 
its  subtending  chord  is  3  inches.  Compute  the  diameter  of 
the  circle. 

566.  The  external  segments  of  two  secants  to  a  circle  from 
the  same  point  are  3.5  m.  and  25  dm.  while  the  internal 
segment  of  the  former  is  150  cm.  What  is  the  internal  seg- 
ment of  the  latter? 

667.  The  hypothenuse  of  a  right  triangle  is  16  feet,  and  the 
perpendicular  to  it  from  the  vertex  jof  the  opposite  angle  is 
5  feet.  Compute  the  values  of  the  legs  and  the  segments  of 
the  hypothenuse. 

568.  The  sides  of  a  certain  triangle  are  6,  7,  and  8  hm. 
respectively.  In  a  similar  triangle  the  side  corresponding  to  8 
is  40  hm.     Compute  the  other  two  sides. 

669.  The  sides  of  a  certain  triangle  are  9,  12,  and  15  feet 
respectively.  Compute  the  segments  of  the  sides  made  by  the 
bisectors  of  the  several  angles. 

670.  If  a  vertical  rod,  4  m.  high,  cast  a  shadow  150  cm.  long, 
how  high  is  a  tree  which,  at  the  same  time  and  place,  casts  a 
shadow  3.5  dkm.  long. 

571.  The  perimeters  of  two  similar  polygons  are  200  and 
300  feet  respectively,  and  one  side  of  the  former  is  24  feet. 
What  is  the  corresponding  side  of  the  latter  ? 


Problems  of  CompiitatioA;    ;>'     J'o^'. 

572.  How  long  must  a  ladder  be  to  reach  a  window  8  m. 
high,  if  the  lower  end  of  the  ladder  is  35  dm.  from  the  side 
of  the  house  ? 

573.  Compute  the  lengths  of  the  longest  and  the  shortest 
chord  that  can  be  drawn  through  a  point  6  inches  from  the 
center  of  a  circle  whose  radius  is  10  inches. 

574.  The  distance  from  the  center  of  a  circle  to  a  chord 
10  dm.  long  is  12  dm.  Compute  the  distance  from  the  center 
to  a  chord  14  dm.  long. 

575.  The  radius  of  a  circle  is  5  inches.  Through  a  point  3 
inches  from  the  center  a  diameter  is  drawn,  and  also  a  chord 
perpendicular  to  the  diameter.  Compute  the  length  of  this 
chord,  and  the  distance  from  one  end  of  the  chord  to  the  ends 
of  the  diameter. 

576.  Through  a  point  55  dm.  from  the  center  of  a  circle  whose 
radius  is  3.5  m.  tangents  are  drawn.  Compute  the  lengths  of 
the  tangents  and  of  the  chord  joining  the  points  of  contact. 

577.  If  a  chord  8  feet  long  be  3  feet  from  the  center  of  the 
circle,  compute  the  radius  and  the  distances  from  the  end  of 
the  chord  to  the  ends  of  the  diameter  which  bisects  the  chord. 

578.  Through  a  point  500  mm.  from  the  center  of  the  circle 
whose  radius  is  13  dm.  any  chord  is  drawn.  What  is  the 
product  of  the  two  segments  of  the  chord?  What  is  the 
length  of  the  shortest  chord  that  can  be  drawn  through  that 
point  ? 

579.  Prom  the  end  of  a  tangent  20  inches  long  a  secant  is 
drawn  through  the  center  of  a  circle.  If  the  exterior  segment 
of  this  secant  be  8  inches,  what  is  the  radius  of  the  circle  ? 

580.  A  tangent  7.2  dkm.  long  is  drawn  to  a  circle  whose 
radius  is  90  dm.  Compute  the  exterior  segment  of  a  secant 
through  the  center  from  the  extremity  of  the  tangent. 


102  Plane  Geometry. 

581.  The  span  of  a  roof  is  28  feet,  and  each  of  its  slopes 
measures  17  feet  from  the  ridge  to  the  eaves.  Compute  the 
height  of  the  ridge  above  the  eaves. 

582.  A  ladder  13  m.  long  is  placed  so  as  to  reach  a  window 
8  m.  high  on  one  side  of  the  street,  and  on  turning  the  ladder 
over  to  the  other  side  of  the  street  it  just  reaches  a  window 
10  m.  high.     What  is  the  width  of  the  street  ? 

583.  The  bottom  of  a  ladder  is  placed  at  a  point  14  feet 
from  a  house,  while  its  top  rests  against  the  house  48  feet 
from  the  ground.  On  turning  the  ladder  over  to  the  other  side 
of  the  street,  its  top  rests  40  feet  from  the  ground.  Compute 
the  width  of  the  street. 

584.  One  leg  of  a  right  triangle  is  3925  feet,  and  the  differ- 
ence between  the  hypothenuse  and  other  leg  is  625  feet.  Com- 
pute the  hypothenuse  and  the  other  leg. 

585.  Compute  the  area  of  a  right  isosceles  triangle  if  the 
hypothenuse  is  100  meters. 

586.  Compute  the  area  of  a  rhombus  if  the  sum  of  its  diago- 
nals is  12  dm.  and  their  ratio  is  3  :  5. 

587.  Compute  the  area  of  a  right  triangle  whose  hypothenuse 
is  13  feet  and  one  of  whose  legs  is  5  feet. 

588.  Compute  the  area  of  an  equilateral  triangle,  one  of 
whose  sides  is  40  hkm. 

689.  The  area  of  a  trapezoid  is  3|-  acres ;  the  sum  of  the  two 
parallel  sides  is  242  yards.  Compute  the  perpendicular  dis- 
tance between  them. 

590.  The  diagonals  of  a  rhombus  are  16.5  m.  and  300  dm. 

respectively.     Compute  its  area. 

591.  The  diagonals  of  a  rhombus  are  88  feet  and  234  feet 
respectively.     Compute  its  area,  and  find  length  of  one  side. 

592.  The  area  of  a  rhombus  is  354,144  sq.  meters,  and  one 
diagonal  is  .672  km.    Compute  the  other  diagonal  and  one  side. 


Problems  of  Computation.  103 

593.  The  sides  of  a  right  triangle  are  in  the  ratio  of  3,  4, 
and  5,  and  the  altitude  upon  the  hypothenuse  is  20  yards. 
Compute  the  area. 

594.  Compute  the  area  of  a  quadrilateral  circumscribed 
about  a  circle  whose  radius  is  25  m.  and  the  perimeter  of  the 
quadrilateral  is  400  m. 

595.  Compute  the  area  of  a  regular  hexagon  inscribed  in  a 
circle  whose  radius  is  12  cm. 

596.  The  base  of  a  triangle  is  75  rods  and  its  altitude  60 
rods.  Compute  the  perimeter  of  an  equivalent  rhombus  if  its 
altitude  is  45  rods. 

597.  Upon  the  diagonal  of  a  rectangle  4.5  m.  by  25  dm.  an 
equivalent  triangle  is  constructed.     Compute  its  altitude. 

598.  Compute  the  side  of  a  square  equivalent  to  a  trapezoid 
whose  bases  are  56  feet  and  44  feet  respectively,  and  each  of 
whose  legs  is  10  feet. 

599.  Find  what  part  of  the  entire  area  of  a  parallelogram 
will  be  the  area  of  the  triangle  formed  by  drawing  a  line  from 
one  vertex  to  the  middle  point  of  one  of  the  opposite  sides. 

600.  In  two  similar  polygons  two  homologous  sides  are  15 
feet  and  25  feet  respectively.  The  area  of  the  first  polygon  is 
450  square  feet.     Compute  the  area  of  the  other  polygon. 

601.  The  base  of  a  triangle  is  32  km.  and  its  altitude  is 
20  km.  Compute  the  area  of  the  triangle  formed  by  drawing 
a  line  parallel  to  the  base  at  a  distance  of  500  dkm.  from  the 
base. 

602.  The  sides  of  two  equilateral  triangles  are  20  and  30 
yards  respectively.  Compute  the  side  of  an  equilateral  tri- 
angle equivalent  to  their  sum. 

603.  If  the  side  of  one  equilateral  triangle  is  equivalent  to 
the  altitude  of  another,  what  is  the  ratio  of  their  areas  ? 


I04  Plane  Geometry. 

604.  The  radius  of  a  circle  is  15  feet,  and  through  a  point 
9  inches  from  the  center  any  chord  is  drawn.  What  is  the 
product  of  the  two  segments  of  this  chord? 

605.  A  square  field  contains  302,500  sq.  m.  Compute  the 
length  of  fence  that  incloses  it. 

606.  A  square  field  210  yards  wide  has  a  path  around  the 
inside  of  its  perimeter  which  occupies  just  one  seventh  of  the 
whole  field.     Compute  the  width  of  the  path. 

607.  A  street  3^  km.  long  contains  5.6  hectars.  How  wide 
is  the  street  ? 

608.  The  perimeter  of  a  rectangle  is  72  feet,  and  its  length 
is  twice  its  breadth.     What  is  its  area  ? 

609.  A  chain  800  dm.  long  incloses  a  rectangle  15  m.  wide. 
How  much  more  area  would  it  inclose  if  the  figure  were  a 
square  ? 

610.  The  perimeter  of  a  square,  and  also  of  a  rectangle 
whose  length  is  four  times  its  breadth,  is  400  yards.  Com- 
pute the  difference  in  their  areas. 

611.  A  rectangle  whose  length  is  25  m.  is  equivalent  to  a 
square  whose  side  is  15  m.  Which  has  the  greater  perimeter, 
and  how  much  ? 

612.  The  perimeters  of  two  rectangular  lots  are  102  yards 
and  108  yards  respectively.  The  first  lot  is  eight  ninths  as 
wide  as  it  is  long,  and  the  second  lot  is  twice  as  long  as  it 
is  wide.  Compute  the  difference  in  the  value  of  the  two  lots 
at  ^  1  per  square  foot. 

613.  A  rhombus  and  a  rectangle  have  equal  bases  and  equal 
areas.  Compute  their  perimeters  if  one  side  of  the.  rhombus 
is  300  m.  and  the  altitude  of  the  rectangle  is  22  dkm. 

614.  The  altitudes  of  two  triangles  are  equal,  and  their 
bases  are  20  feet  and  30  feet  respectively.  Compute  the  base 
of  a  triangle  equivalent  to  their  sum  and  having  an  altitude 
one  fourth  as  great. 


Measurement  of  the  Circle.  105 


MEASUREMENT  OF  THE  CIRCLE.  — THEOREMS. 

615.  If  radii  of  a  regular  circumscribed  polygon  of  22  sides  be  drawn, 
the  chords  joining  successive  points  of  intersection  will  form  a  regular 
polygon  of  n  sides,  and  the  sides  of  the  two  polygons  will  be  parallel 
two  and  two. 

616.  If  tangents  to  a  circle  be  drawn  parallel  to  the  sides  of  a  reg- 
ular inscribed  polygon  of  22  sides,  they  will  form  a  regular  polygon  of 
22  sides. 

617.  If,  in  the  same  circle,  a  circumscribed  and  inscribed  polygon 
of  n  sides  have  their  sides  parallel,  two  and  two,  their  ra-dii  are  con- 
gruent. 

618.  If  two  regular  polygons  of  the  same  number  of  sides  be  in- 
scribed in,  or  circumscribed  about,  the  same  circle,  they  are  equal. 

619.  If  tangents  are  drawn  at  the  middle  points  of  the  arcs  and 
terminating  in  the  sides  of  a  regular  circumscribed  polygon  of  n  sides, 
a  regular  circumscribed  polygon  of  2  12  sides  will  thus  be  formed, 

620.  If  the  vertices  of  a  regular  inscribed  polygon  of  22  sides  are 
joined  to  the  middle  points  of  the  arcs  subtended  by  the  sides  of  the 
polygon,  the  lines  thus  drawn  will  form  a  regular  inscribed  polygon  of 
2  22  sides. 

621.  The  perimeter  of  a  regular  inscribed  polygon  of  22  sides  is  less 
than  the  perimeter  of  the  regular  polygon  of  2  22  sides  inscribed  in  the 
same  circle. 

622.  The  perimeter  of  the  regular  circumscribed  polygon  of  22  sides 
is  greater  than  the  perimeter  of  the  regular  polygon  of  2  22  sides  cir- 
cumscribed about  the  same  circle. 

623.  If  the  extremities  of  a  line  be  joined  by  the  arc  of  a  circle  and 
by  other  lines  which  envelop  this  arc,  the  arc  is  the  least  of  them  all. 

Post.  Let  AB  be  a  chord  of  the  circle  ABQ,  subtending  the 
arc  ANB,  and  let  AOB  be  any  one  of  the  lines  enveloping  the 
arc  ANB. 


io6 


Plane  Geometry. 


To  Prove.    Arc  ANB  is  the  least  of  them  all. 

Dem.    AOB  envelops  ANB.  (?) 

.*.  none  of  its  points  are  within  the 

circle,   and  all   its  points   cannot   be 

on  ANB ;  i.e.  some  of  its  points  must 

be  outside  ANB. 

.'.  from  some  point  in  ANB,  as  N,  a 

tangent  can  be  drawn  terminating  in 

^0^,  asati^and^. 

FH<FOH  (?) 

.-.  AFNHB  <  AFOHB         (?) 

.-.  AOB  is  not  the  least  line  joining  A  and  B.     In  like 

manner  it  can  be  shown  that  no  enveloping  line  can  be  the  least. 

.-.  ANB  is  the  least.  q.e.d. 

624.  If  two  circles  with  unequal  radii  intersect  each  other, 
of  the  lesser  arcs  subtended  by  their  common  chord,  the  arc  of 
the  larger  circle  is  less  than  the  arc  of  the  smaller  circle. 


Post.   Let  DCII  and  DCF  be  two  intersecting  circles  of 
which  DCHis  the  larger,  and  having  the  common  chord  CD. 

To  Prove.  Arc  CND  <  arc  COD. 


Measurement  of  the  Circle. 


107 


Cons.  Join  their  centers  by  line  AB.  Draw  tlie  radii  AC 
and  BC.  Draw  CQ  =  CB  and  with  Q  as  a  center  and  radius 
QC  construct  circle  DCR. 

Bern.  Aq>AC-  QC  Why  ? 

.*.  circumference  CDR  will  intersect  QJ5  between  N  and  F  as 
at  P. 

li  AQ  =  AC  —  QC,  they  would  be  tangent  to  each  other,  and 
if  AQ  <  AC  —  QC,  one  lies  within  the  other,  and  both  these 
results  contradict  the  Hyp. 

arc  CPD  =  arc  COD  Why  ? 

arc  CND  <  arc  CPD  Why  ? 

.-.  arc  CND  <  arc  COD  Why?     q.e.d. 

625.  If  a  circle  be  conceived  to  roll  along  a  straight  line 
without  slipping  and  keeping  always  in  the  same  plane,  a 
point  in  the  circumference  will  describe  a  curve  called  a 
cycloid,  e.g. 


Suppose  the  circle  DOP,  which  is  tangent  to  line  AB  at  D, 
to  roll  along  the  line  AB  without  slipping,  in  the  direction  of 
the  arrow  and  keeping  always  in  the  same  plane.  The  center 
B  will  move  in  the  line  FII  parallel  to  AB,  while  the  point  D 
will  describe  the  curve  DKC,  which  is  called  the  cycloid. 

It  is  evident  from  the  definition 

i.  That  the  line  CD  is  equal  in  length  to  the  circumference 
of  the  circle. 

ii.   That  the  line  CiVis  equal  in  length  to  the  arc  QN. 


io8 


Plane  Geometry. 


If,  from  a  point  without  a  circle,  two  tangents  are 
drawn,  the  lesser  of  the  two  arcs 
joining  the  points  of  contact  is 
less  than  the  sum  of  the  two 
tangents. 

Post.  Let  CBN  be  a  circle, 
H  its  center,  and  AC  and  AB 
two  tangents  to  that  circle 
drawn  from  point  A,  the  points 
of  contact  being  C  and  B. 

To  Prove.     That  arc   CB  is 
less  than  AC  +  AB. 
Cons.     Suppose  AH  to  be  drawn,  and  let  the  curve  DFN  be 
the  cycloid  traced  by  the  point  F  as  the  circle  rolls  along  the 
line  OA  toward  A, 


Dem. 

DB<AB         Why? 

.-.  Arc 

CF<AC         Why? 

Arc 

FB  =  DB       See  625. 

.-.  Arc 

FB  +  arc  CF<  AC 

.*.  Arc 

FB  <  AB          Why  ? 

+  AB         Why  ? 

Arc 

FB^SiVG  CF    Why? 

or  Arc 

CB<AC-}-AB 

AC  =  AB          Why? 

Q.E.D. 

627.  The  circumfereiice  of  a  circle  is  greater  than  the  perimeter  of 
any  polygon  inscribed  in  it. 

628.  The  circumference  of  a  circle  is  less  than  the  perimeter  of  any 
polygon  circumscribed  about  it. 

629.  The  area  of  a  regular  inscribed  polygon  of  2  12  sides  is  greater 
than  the  area  of  a  regular  polygon  of  n  sides  inscribed  in  the  same  circle. 

630.  The  area  of  a  regular  circumscribed  polygon  of  2  22  sides  is 
less  than  the  area  of  a  regular  polygon  of  22  sides  circumscribed  about 
the  same  circle. 

631.  If  the  number  of  sides  of  a  regular  inscribed  polygon  be  con- 
tinuously doubled,  its  perimeter  will  become  an  increasing  variable  and 
approach  the  circumference  as  its  limit. 


Measurement  of  the  Circle.  109 

632.  If  the  number  of  sides  of  a  regular  inscribed  polygon  be  con- 
tinuously doubled,  its  apotbem  will  become  an  increasing  variable,  and 
approach  the  radius  as  its  limit. 

633.  If  the  number  of  sides  of  a  regular  circumscribed  polygon  be 
continuously  doubled,  its  perimeter  will  become  a  decreasing  variable, 
and  approach  the  circumference  as  its  limit. 

634.  If  the  number  of  sides  of  a  regular  circumscribed  polygon  be 
continuously  doubled,  its  radius  will  become  a  decreasing  variable,  and 
approach  the  radius  of  the  circle  as  its  limit. 

635.  If  the  number  of  sides  of  a  regular  inscribed  polygon  be  con- 
tinuously doubled,  its  area  will  become  an  increasing  variable  and 
approach  the  area  of  the  circle  as  its  limit. 

636.  If  the  number  of  sides  of  a  regular  circumscribed  polygon  be 
continuously  doubled,  its  area  will  become  a  decreasing  variable  and 
approach  the  area  of  the  circle  as  its  limit. 

637.  The  area  of  a  regular  inscribed  polygon  of  2  22  sides  is  a  mean 
proportional  between  the  areas  of  two  polygons,  each  of  n  sides,  one 
inscribed  within,  and  the  other  circumscribed  about,  the  same  circle. 

Post.  Let  AB  be  one  side  of  a  regular  inscribed  polygon  of 
n  sides,  and  EF  one  side  of  a  regular 
circumscribed  polygon  of  n  sides  and 
parallel  to  AB.  Let  CE  and  CF  be 
radii  of  this  polygon.  Also  let  CG 
be  a  radius  of  the  circle  drawn  to 
point  of  contact.  Let  AH  and  BI  be 
tangents  and  J.(t  a  chord  of  the 
circle.  Let  also  CH  and  CI  be 
drawn.  Designate  the  area  of  the 
polygon  whose  side  is  AB  by  p,  that 
of  the  polygon  whose  side  is  EF  by  P,  and  that  whose  side  is 
AG  by  p\ 

To  Prove.  p:p' :  :p' :  P. 


1 1  o  Plane  Geometry, 

Dem. 


Area.                            Area. 

Area. 

2  n  .  AGD  =i),  2  n  .  ACO=p\  and  2  n 

■  Eca=p. 

Area  ^OD :  area  ACQ  ::CD:CG 

(?) 

Area  J. CD  :  area  ACG  ::p:p' 

(?) 

.:  p:p'::CD:CG 

(?) 

Area  CAG  :  area  C^G^  ::CA:CE 

(?) 

Ave3i  GAG  :Sive3^CEG:p':P 

(?) 

.-.  CA:CE::p':P 

(?) 

CA:CE::CD:CG 

(?) 

.-.  p:p'::p':P 

(?)    Q.E.D. 

638.  If  polygons  of  n  and  2  n  sides  be  inscribed  in,  and  circumscribed 
about,  tbe  same  circle,  the  area  of  the  circumscribed  polygon  of  2  22 
sides  equals  twice  the  product  of  the  areas  of  the  polygons  of  n  sides 
divided  by  the  sum  of  the  areas  of  the  two  inscribed  polygons. 

Post.  Use  same  diagram  as  in  previous  theorem,  and  desig- 
nate areas  the  same,  also  designate  area  of  the  polygon  whose 
side  is  JET/ by  P'. 

To  Prove,  p  ^llL. 

p+p' 

Dem.        Area  CGH :  area  CHE  ..Gil'.  HE  (?) 

CG:CE::GH:HE  (?) 

.-.  Area  CGH:  area  CHE : :  CG  :  CE  (?) 

Prom  the  Dem.  in  637,  3d  proportion 

p:20'::CD:CG 

CG.CE.-.CD'.CA  (?) 

or  CG:CE::CD:CG  (?) 

.'.  p:p'::CG:CE  (?) 

.♦.  Area  CGH :  area  CHE  ::p:p'  (?) 

. ••  Area  CGH :  area  CGE  ::p:p+p'  (?) 

.♦.  2  Area  C(?^:area  CGE  :  :2  p-.p+p'  (?) 

or  Area  CHI :  area  CGE  ::2p:p-}-p'  (?) 


Measurement  of  the  Circle.  1 1 1 

Area.  Area. 

2  n  '  CHI=  P',  and  2  71 .  CGE  =  P  (?) 

.-.  Area  CHI:  area  CGE  iiP'iF  (?) 

.-.  F':P::2p:p+p^  (?) 

.-.    P'  =  ^I^.  (?)     Q.E.D. 

p+p' 

639.  The  chords  which  join  the  extremities  of  two  perpendicular 
diameters  form  a  square. 

640.  The  diagonals  of  an  inscribed  square  are  diameters  of  the  cir- 
cumscribed circle. 

641.  The  tangents  to  a  circle  whose  points  of  contact  are  the  vertices 
of  an  inscribed  square  will  form  a  square. 

642.  The  area  of  an  inscribed  square  is  equal  to  twice  the  square  of 

the  radius. 

643.  The  area  of  a  circumscribed  square  is  equal  to  four  times  the 
square  of  the  radius. 

644.  Problem.  To  compute  the  areas  of  the  inscribed  and  circum- 
scribed polygons  of  2  22  sides,  having  given  the  areas  of  those  of  n  sides. 
Prom  637  and  638  we  have 

I.  p'=Vp^  and  11.  P'  =^^. 

Letting  p  and  P  represent  the  area  of  inscribed  and  circnm- 
scribed  squares  respectively  and  designating  the  radius  by  ?•, 
we  have  from  642  and  643 

p  =  2r2  and  P=4.t^. 

By  substituting  these  values  in  I.,  we  have 

p'  =  2.82843  r". 
Then  by  substituting  values  otp,p'j  and  P  in  II.,  we  have 

P'  =  3.31371  r^. 

Thus  we  have  computed  the  areas  of  the  regular  inscribed  and 
circumscribed  octagons  in  terms  of  the  radius  of  the  circle. 


I  12 


Plane  Geometry. 


Again,  calling  these  p  and  P,  p'  and  P'  will  be  polygons  of 
sixteen  sides ;  and  using  the  f ormulaj  as  before,  the  areas  of  the 
latter  may  be  computed.  Eepeating  the  process,  those  of  thirty- 
two,  sixty-four,  etc.,  sides  may  be,  in  like  manner,  computed. 

Below  are  the  tabulated  results  to  seven  decimal  places  for 
thirteen  doublings  of  the  number  of  sides  of  the  polygons : 


Number  of  Sides. 

Area  of  In- 
scribed Polygon. 

Area  of  Circum- 
scribed Polygon. 

4 

2.0000000  r" 

4.0000000  r" 

8 

2.8284271 1^ 

3.3837085  7^ 

16 

3.0614675  r" 

3.1825979  i^ 

32 

3.1214452  r^ 

3.1517240  7^  • 

64 \     . 

3.1365485  r^ 

3.1441184  r^ 

128 

3.1403312  7^ 

3.1422236  r^ 

2bQ 

3.1412773  7^ 

3.1417504  r^ 

512 

3.1415138  r^ 

3.1416321 1^ 

1024 

3.1415729  r" 

3.1416025  7-2 

2048 

3.1415877  r^ 

3.1415951  r" 

4096 

3.1415914  r" 

3.1415933  r^ 

8192 

3.1415912  r" 

3.1415928  r2 

16384 

3.1415925  r" 

3.1415927  r" 

32768  

3.1415926  r^ 

3.1415926  r^ 

Hence,  since  the  area  of  the  circle  is  greater  than  that  of  the 
inscribed  polygon  and  less  than  that  of  the  circumscribed  poly- 
gon (635  and  636),  3.1415926  r^  must  be  the  area  of  the  circle 
correct  to  within  less  than  one  ten-millionth  part  of  r*.  But  by 
continuing  the  process,  the  areas  of  the  two  polygons  may  be 
made  to  agree  to  any  desired  number  of  decimal  places,  and, 
therefore,  such  result  may  be  taken  as  the  area  of  the  circle 
without  sensible  error.  If  r  be  taken  as  unity,  it  would,  of 
course,  vanish  from  the  expression,  and  consequently  3.1415926 
may  be  taken  as  the  area  of  the  circle  whose  radius  is  unity, 
correct  to  seven  decimal  places. 


Measurement  of  the  Circle, 


"3 


645.   If  in  637  and  638  we  let  P,  P',  p,  and  p'  represent  perimeters 
instead  of  areas,  then, 

I.   P'  =  ^^,  and  II.  p'=VpxT'. 
p-\-F 

Dem.  and 

P:p::EC:AC(ovCGf)  (?) 

EH:HG::EC:CG  (?) 

.'.P:p::EH:HG  (?) 
.-.  P+p'.2p'.'.EH 

+  HG'.2HG  (?) 

.'.P+P'.2p'.'.EG',HI  (?) 
But 

P:P'::EG:HI  (?) 

/.  P+p:2p::P:P'  (?) 

i>  +  P  ^^ 

Again 

AD:AG::p:p'  (?) 


AG:HI::p':P 

(?) 

.  AGHI,,,p, 
"    2    '  2   "^  ' 

(?) 

OG:HG::AD:AG 

(?) 

OG  =  ^^,^ndHG  = 

2   ^  '' 

.■.f:f::AD..Aa 

(?) 

.-.AD-.AG-.-.p'-.P' 

(?) 

.■.p:p'::p':P' 

(?) 

.■.p'=^p-P< 

(?) 

Q.E.D. 


646.  We  will  now  use  these  formulsB  for  the  computation  of  perim- 
eters in  a  manner  similar  to  that  in  644,  beginning  with  the  circiim- 
scribed  and  inscribed  squares. 

The  perimeter  of  the  circumscribed  square  in  terms  of  the 
diameter,  calling  the  latter  D,  is, 
obviously,  4  2).     In  the  right  tri- 
angle APD 

Aff  =  1^  +  ^0^  =  21^  =  21^ 


-(t)'=f 


,'.AD=^,   4.AD=—D=I>VS. 

V2  V2 

Hence  the  perimeter  of  the  in- 


K 


D 


G 


If 


114 


Plane  Geometry. 


scribed  square  is  2.8284271  Z).     Continuing  as  in  644  and  tabu- 
lating the  results,  we  have  the  following: 


Number  of  Sides. 

Perimeter  of 
Inscribed  Polygon. 

Perimeter  of  Cir- 
cumscribed Polygon. 

4 

8  .  . 

16 

32 

64 

128 

256 

512 

1024  .  

2048 

4096 

8192 

16384   

32768   

2.8284271 D 
3.0614675  Z) 
3.1214452  Z) 
3.1365485  Z) 
3.1403312  Z) 
3.1412773  Z) 
3.1415138  Z) 
3.1415729  Z) 
3.1415877  Z) 
3.1415914  Z) 

3.1415923  Z) 

3.1415924  Z) 

3.1415925  Z) 

3.1415926  Z) 

4.0000000  Z) 
3.3137085  Z) 
3.1825979  Z) 
3.1517249  Z) 
3.1441184  Z) 
3.1422236  Z) 
3.1417504  Z) 
3.1416321  D 
3.1416025  Z) 
3.1415951  Z) 
3.1415933  Z) 
3.1415928  Z) 
3.1415927  Z) 
3.1415926  Z) 

Hence,  since  the  circumference  of  the  circle  is  greater  than 
the  perimeter  of  the  inscribed  polygon,  and  less  than  that  of  the 
circumscribed  polygon  (627  and  628),  3.1415926  Z)  must  be  the 
circumference  of  the  circle  correct  to  within  less  than  one  ten- 
millionth  part  of  Z).  But  by  continuing  the  process  the  perim- 
eters of  the  two  polygons  may  be  made  to  agree  to  any  desired 
number  of  decimal  places,  and  therefore  such  result  may  be 
taken  as  the  circumference  of  the  circle  without  sensible  error. 
If  Z)  be  taken  as  unity,  it  would,  of  course,  vanish  from  the 
expression,  and  consequently  3.1415926  may  be  taken  as  the 
circumference  of  a  circle  whose  diameter  is  unity,  correct  to 
seven  decimal  places. 

647.  The  circumferences  of  two  circles  are  in  the  same  ratio  as  their 
radii,  and  also  their  diameters. 


Measurement  of  the  Circle, 


"5 


Post.  Let  ABHKj  etc.,  and  STVY,  etc.,  be  two  circles 
whose  centers  are  N  and  TF,  and  designate  the  circumferences 
by  C  and  c,  and  their  radii  by  R  and  r,  and  their  diameters  by 
D  and  d,  the  capital  letter  referring  to  the  larger  circle. 

^To  Prove,  1.  C:c::  R:r 

II.   C:c::D:d 

Cons.    Inscribe  in  each  a  regular  polygon  of  n  sides,  and 
construct  the  radii  NB  and  WT,  and  the  apothems  NQ  and  WE. 
Dem.    Designating  the  perimeters  by  P  and  p, 

P'.p'.-.NB'.WT  (?) 

.-.  P:^5::p:TFr  (?) 

or  ^=.JP-.  (?) 

NB      WT  ^  ^ 

We  will  now  inscribe  polygons  with  double  the  number  of 
sides,  and  continue  this  process  indefinitely ;  then  P  and  p  be- 
come increasing  variables  approaching  C  and  c  as  their  limits. 


P 
NB 


becomes  an  incr.  var.  appr. 


its  limit  and  -~— 

NB  WT 


becomes  an  incr.  var.  appr.  its  limit • 

^^  WT 


or  —  = 


%.e. 


NB     WT'       R 
C:R::c:r 
.'.  C:c::R:r 
.'.  C'.ci.Did 


(?) 

(?) 
(?) 


Q.E.D. 


1 1 6  Plane  Geometry. 

648.  The  areas  of  two  circles  are  in  tte  same  ratio  as  tlie  squares 
of  their  radii  and  of  their  diameters. 

Sug.     Use  method  similar  to  the  above,  and  consult  540. 

649.  Similar  arcs  are  in  the  same  ratio  as  the  radii  of  the  circum- 
ferences of  which  they  are  a  part,  an'd  also  as  the  diameters. 


Post.     Let  CB  and  KH  be  two  similar  arcs,  and  A  and  D 
the  centers  of  the  circumferences  of  w^hich  they  are  a  part. 
Cons.     Draw  the  radii  AC,  AB,  DK,  and  DH. 
(Designate  circumferences,  diameters,  and  radii,  as  before.) 

To  Prove.        I.  Arc  CB :  arc  KH: :  R:r 

11.  Arc  CB:2iYcKH::D:d 

Dem.     ZA=:ZD.     Why? 
Hence  arc  CB  is  the  same  part  of  the  circumference  0  as  arc 
KH  is  of  the  circumference  c. 

.'.  Arc  CB  :  arc  KH:  :C:c  (?) 

But                                                  R:r::C:c  (?) 

.-.  I.  Arc  CB  :  arc  KH:  :B:r  (?) 

11.  Arc  CB  :  arc  KH:  :D:d  q.e.d. 

650.  The  areas  of  similar  sectors  are  in  the  same  ratio  as  the  squares 
of  the  radii,  and  also  of  the  diameters,  of  the  circles  of  which  they  are 
a  part. 

651.  The  area  of  a  circle  is  equal  to  one  half  the  product  of  its  cir- 
cumference and  radius. 

S2ig.     Consult  541. 

652.  The  area  of  a  sector  is  equal  to  one  half  the  product  of  its  arc 
and  radius. 


Measurement  of  the  Circle.  117 

X- 

653.  The  areas  of  similar  segments  are  in  the  same  ratio  as  the 
squares  of  their  radii,  the  squares  of  their  diameters,  and  as  the  squares 
of  their  chords. 

Sug.  Consult  Theorems  650  and  533  and  utilize  the  fact 
that  the  area  of  the  segment  is  equal  to  the  area  of  the  sector 
minus  the  area  of  the  triangle.  For  brevity,  designate  the 
areas  of  the  triangles  by  T  and  t,  the  areas  of  the  sectors  by 
JTand  k,  and  the  areas  of  the  segments  by  aS'  and  s. 

654.  I.  Let  us  designate  the  circumference  of  a  circle  whose  diam- 
eter is  unity  by  tt,  and  the  circumference  of  any  other  circle  by  (7; 
its  diameter  by  D ;  its  radius  by  U ;  and  its  area  by  A. 

Then  C.iT'.'.D'.l.  (?) 

Hence  I.  C  —  ttD, 

Q 

whence  11.  ^  =  'r, 

or  III.  C=w2B. 

■n 

Multiplying  both  members  of  this  equation  by  —,  we  have, 

OR 

But,  by  651,  ^  =  the  area  of  the  circle ;  hence 

IV.  A  =  7rE\ 

Hence  the  area  of  any  circle  is  equal  to  the  square  of  its 
radius  multiplied  by  the  constant  quantity  tt,  and  the  circum- 
ference of  every  circle  is  equal  to  the  product  of  its  diameter 
(or  twice  its  radius)  by  the  same  quantity  tt. 

From  II.  above,  it  is  readily  seen  that  tt  is  the  ratio  of  the 
circumference  of  any  circle  to  its  diameter,  or  of  a  semicir- 
cumference  to  its  radius. 

The  exact  numerical  value  of  tt  can  be  only  approximately 
expressed.      As  computed  in  646  it  is  3.1415926,   but   for 


1 1 8  Plane  Geometry. 

practical  purposes  in  computing  its  value  is  usually  taken  as 
3.1416.     In  many  cases  3.14,  or  '^-f,  is  sufficiently  accurate. 

The  symbol  tt  is  the  first  letter  of  the  Greek  word  meaning 
perimeter,  or  circumference. 

II.  The  quadrature  of  the  circle  is  the  problem  which  requires 
the  finding  of  a  square  which  shall  be  equal  in  area  to  that 
of  a  circle  with  a  given  radius.  Now  since  the  area  of  a  circle 
is  equal  to  its  circumference  multiplied  by  one  half  its  radius, 
if  a  straight  line  of  same  length  as  circumference  be  taken  as 
the  base  of  a  rectangle,  and  one  half  the  radius  as  its  altitude, 
their  product  will  be  the  area  of  the  rectangle,  also  of  the 
circle.  It  is  also  evident  that,  if  a  line  which  is  a  mean  pro- 
portional between  these  two  be  taken  as  the  side  of  a  square, 
the  area  of  this  square  will  be  equal  to  that  of  the  rectangle, 
and  consequently  to  that  of  the  circle.  It  will  be  shown  in 
the  series  of  problems  of  construction  (Prob.  837)  how  this 
mean  proportional  can  be  found ;  hence,  to  "  square  the  circle  " 
we  must  be  able  to  find  the  circumference  when  the  radius  is 
known,  or  vice  versa.  For  accomplishing  this,  we  must  know 
the  ratio  of  the  circumference  to  its  diameter  or  radius.  But 
this  ratio,  as  has  been  remarked  before,  can  be  only  approxi- 
mately expressed;  for,  as  the  higher  mathematics  prove,  the 
circumference  and  diameter  are  incommensurable,  but  the 
approximation  has  been  carried  so  far  that  the  error  is  infini- 
tesimal. Archimedes,  about  250  B.C.,  was  the  first  to  assign 
an  approximate  value  to  tt.  He  found  that  it  must  be  between 
3.1428  and  3.1408.  In  1540  Metius  computed  it  correctly  to 
6  places.  Later,  in  1579,  Vieta  carried  the  approximation  to 
10  places.  Van  Ceulen  to  36  places.  Sharp  to  72  places,  Machin 
to  100  places,  De  Lagny  to  128  places,  Rutherford  to  208 
places,  and  Dr.  Clausen  to  250  places.  In  1853  Eutherford 
carried  it  to  440  places,  and  in  1873  Shanks  computed  it  to 
707  places,  but  these  latter  results  do  not  appear  to  have  been 
verified.  The  following  is  its  value  to  208  places,  as  computed 
by   Eutherford.     It   should  be   stated,   however,   that   these 


Measurement  of  the  Circle.  1 1 9 

extensive  computations  are  by  the  more  expeditious  methods 
of  trigonometry. 

TT  =  3.141592653589793238462643383279- 
502884197169399375105820974944- 
592307816406286208998628034825- 
342717067982148086513282306647- 
093844609550582231725359408128- 
484737813920386338302157473996- 
0082593125912940183280651744+ 

Some  idea  of  the  accuracy  of  the  above  value  may  be  formed 
from  the  following  statement  taken  from  Peacock's  Calculus : 
"If  the  diameter  of  the  universe  be  100,000,000,000  times  the 
distance  of  the  sun  from  the  earth  (about  93,000,000  miles), 
and  if  a  distance  which  is  100,000,000,000  times  this  diameter 
be  divided  into  parts,  each  of  which  is  one  100,000,000,000th 
part  of  an  inch ;  then  if  a  circle  be  described  whose  diameter 
is  100,000,000,000  times  as  often  as  each  of  those  parts  of  an 
inch  is  contained  in  it ;  then  the  error  in  the  circumference  of 
this  circle,  as  computed  from  this  approximation,  will  be  less 
than  100,000,000,000th  part  of  the  one  100,000,000,000th  part 
of  an  inch." 

ADVANCE    THEOREMS. 

655.  In  two  circles  of  different  radii,  angles  at  the  center 
subtending  arcs  of  equal  lengths  are  to  each  other  inversely  as 
the  radii. 

656.  If,  from  any  point  within  a  regular  polygon  of  n  sides, 
perpendiculars  be  drawn  to  all  the  sides,  the  sum  of  these 
perpendiculars  is  equal  to  n  times  the  apothem. 

657.  An  equiangular  polygon  inscribed  in  a  circle  is  regular 
if  the  number  of  its  sides  be  odd. 

658.  An  equilateral  polygon  circumscribed  about  a  circle  is 
regular  if  the  number  of  its  sides  is  odd. 


I20  Plane  Geometry. 

659.  The  sum  of  the  squares  of  the  lines  joining  any  point 
in  the  circumference  of  a  circle  with  the  vertices  of  an  inscribed 
square,  is  equal  to  twice  the  square  of  the  diameter  of  the 
circle. 

660.  The  area  of  the  ring  included  between  two  concentric 
circles  is  equal  to  the  area  of  the  circle  whose  diameter  is  that 
chord  of  the  outer  circle  which  is  tangent  to  the  inner. 

661.  If  three  circles  be  described  upon  the  sides  of  a  right 
triangle  as  diameters,  the  area  of  that  described  upon  the 
hypothenuse  is  equal  to  the  sum  of  the  areas  of  the  other  two. 

662.  If  upon  the  legs  of  a  right  triangle  semicircumferences 
are  described  outwardly,  the  sum  of  the  areas  contained  between 
these  semicircumferences  and  the  semicircumference  passing 
through  the  three  vertices  is  equal  to  the  area  of  the  triangle. 

663.  If  a  semicircle  be  described  on  the  chord  of  a  quadrant 
as  a  diameter,  the  area  of  the  crescent  is  equal  to  that  of  the 
triangle  of  the  quadrant. 

664.  If  the  diameter  of  a  circle  be  divided  into  two  parts, 
and  upon  these  parts  semicircumferences  are  described  on 
opposite  sides  of  the  diameter,  these  semicircumferences  will 
divide  the  circle  into  two  parts,  which  have  the  same  ratio  as 
the  two  parts  of  the  diameter. 

665.  If  two  chords  of  a  circle  be  perpendicular  to  each 
other,  the  sum  of  the  areas  of  the  four  circles  described  upon 
the  four  segments  as  diameters  will  be  equal  to  the  area  of  the 
given  circle. 

666.  If  squares  be  constructed  outwardly  upon  the  sides  of 
a  regular  hexagon,  their  exterior  vertices  will  be  the  vertices 
of  a  regular  dodecagon. 

667.  The  radius  of  a  regular  inscribed  polygon  is  a  mean 
proportional  between  its  apothem  and  the  radius  of  a  regular 
polygon  of  the  same  number  of  sides  circumscribed  about  the 
same  circle. 


Problems  of  Computation.  121 

668.  The  area  of  a  regular  dodecagon  is  equal  to  three  times 
the  square  of  its  radius. 

669.  If  the  radius  of  a  circle  be  divided  in  extreme  and 
mean  ratio,  the  larger  part  will  be  equal  to  the  side  of  a 
regular  decagon  inscribed  in  the  same  circle. 

670.  The  apothem  of  a  regular  inscribed  pentagon  is  equal 
to  one  half  the  sum  of  the  radius  of  the  circle  and  the  side 
of  the  regular  decagon  inscribed  in  the  same  circle. 

671.  The  square  of  the  side  of  a  regular  inscribed  pentagon 
is  equivalent  to  the  sum  of  the  squares  of  the  side  of  the 
regular  inscribed  decagon  and  the  radius  of  the  circle. 

PROBLEMS    OF    COMPUTATION. 

672.  Compute  the  side  of  a  regular  circumscribed  trigon  in 
terms  of  the  side  of  the  regular  trigon  inscribed  in  the  same 
circle.     Compare  their  areas. 

673.  Compute  the  side  of  a  circumscribed  square  in  terms 
of  the  side  of  the  square  inscribed  in  the  same  circle. 

674.  Compute  the  apothem  of  a  regular  inscribed  trigon  in 
terms  of  the  side  of  the  regular  hexagon  inscribed  in  the  same 
circle. 

675.  Compute  the  apothem  of  a  regular  inscribed  hexagon  in 
terms  of  the  side  of  the  regular  trigon  inscribed  in  the  same 
circle. 

676.  Eegular  trigons  and  hexagons  are  both  inscribed  in,  and 
circumscribed  about,  the  same  circle.     Compare  their  areas. 

677.  Compute  the  area  of  a  regular  polygon  of  12  sides  in- 
scribed in  a  circle  whose  radius  is  50  cm. 

678.  Compute  the  perimeter  of  a  regular  pentagon  inscribed 
in  a  circle  whose  radius  is  10  feet. 

679.  The  perimeter  of  a  regular  hexagon  is  480  m.,  and  that 
of  a  regular  octagon  is  the  same.  Which  has  the  greater  area, 
and  how  much  ? 


122  Plane  Geometry, 


680.  If  paving  blocks  -are  in  the  shape  of  regular  polygons 
(i.e.  their  cross-sections),  how  many  shapes  can  be  employed  in 
order  to  completely  fill  the  space  ? 

681.  Compute  the  diameter  of  a  circle  whose  circumference 
is  225  dm. 

682.  The  diameter  of  a  carriage  wheel  is  4  feet  and  3  inches. 
How  many  revolutions  does  it  make  in  traversing  one  fourth  of 
a  mile  ? 

683.  What  is  the  width  of  a  ring  between  two  concentric 
circumferences  whose  lengths  are  480  cm.  and  3.6  m.  ? 

684.  Find  the  length  of  an  arc  of  36°  in  a  circle  whose 
diameter  is  36  inches. 

685.  In  raising  water  from  the  bottom  of  a  well  by  means  of 
a  wheel  and  axle  it  was  found  that  the  axle,  whose  diameter 
was  20  cm.,  made  20  revolutions  in  raising  the  bucket.  Com- 
pute the  depth  of  the  well. 

686.  Find  the  central  angle  subtending  an  arc  6  feet  and 
4  inches  long,  if  the  radius  of  the  circle  be  8  feet  and  2  inches. 

687.  If  the  radius  of  a  circle  is  6  meters,  find  the  perimeter 
of  a  sector  whose  angle  is  45°. 

688.  If  the  central  angle  subtending  an  arc  10  feet  and  6 
inches  long  is  72°,  what  is  the  length  of  the  radius  of  the  circle  ? 

689.  If  the  length  of  the  meridian  of  the  earth  be  40,000,000 
meters,  what  is  the  length  of  an  arc  of  1"? 

690.  Two  arcs  have  the  same  angular  measure,  but  the  length 
of  one  is  twice  that  of  the  other.  Compare  the  radii  of  those 
arcs. 

691.  Compute  the  area  of  a  circle  whose  circumference  is 
100  dkm. 

692.  Two  arcs  have  the  same  length,  but  their  angular 
measurements  are  20°  and  30°  respectively.  If  the  radius  of 
the  first  arc  is  6  feet,  compute  the  radius  of  the  other. 


Problems  of  Computation.  123 

693.  Find  the  circumference  of  a  circle  in  meters  whose  area 
is  2.5  hektares. 

694.  The  diameter  of  a  circle  is  40  feet.  Find  the  side  of  a 
square  which  is  double  the  area  of  the  circle. 

695.  The  area  of  a  square  is  196  square  meters.  Find  the 
area  of  a  circle  inscribed  in  the  square. 

696.  A  circular  fish-pond  which  covers  an  area  of  5  acres 
and  100  square  rods  is  surrounded  by  a  walk  5  yards  wide. 
Compute  the  cost  of  graveling  the  walk  at  6J  cents  per  square 
yard. 

697.  What  must  be  the  width  of  a  walk  around  a  circular 
garden  containing  If  hektares,  in  order  that  the  walk  may  con- 
tain exactly  one  fourth  of  a  hektare  ? 

698.  A  carpenter  has  a  rectangular  piece  of  board  15  inches 
wide  and  20  inches  long,  from  which  he  wishes  to  cut  the 
largest  possible  circle.  How  many  square  inches  of  the  board 
must  he  cut  away  ? 

699.  The  perimeter  of  a  circle,  a  square,  and  a  regular  trigon 
are  each  equal  to  144  kilometers.     Compare  their  areas. 

700.  If  the  radius  of  a  circle  be  12  inches,  what  is  the  radius 
of  a  circle  10  times  as  large  ? 

701.  What  will  it  cost  to  pave  a  circular  court  240  dkm.  in 
diameter  at  $  4.75  per  square  meter,  leaving  in  the  center  a 
hexagonal  space,  each  side  of  which  measures  2.5  meters. 

702.  A  circle  18  feet  in  diameter  is  divided  into  three 
equivalent  parts  by  two  concentric  circumferences.  Find  the 
radii  of  these  circumferences. 

703.  If  the  chord  of  an  arc  be  720  dm.,  and  the  chord  of  its 
half  be  38  meters,  compute  the  diameter  of  the  circle. 

704.  The  chord  of  half  an  arc  is  17  feet,  and  the  height  of 
the  arc  is  7  feet.     Compute  the  diameter  of  the  circle. 


124  Plane  Geometry. 

705.  The  lengths  of  two  chords,  drawn  from  the  same  point 
in  the  circumference  of  a  circle  to  the  extremities  of  a  diameter, 
are  6  feet  and  8  feet  respectively.  Compute  the  area  of  the 
circle. 

706.  The  area  of  a  sector  is  16.5  ares,  and  the  angle  of  the 
sector  is  36°.  Compute  the  radius  of  the  circle  and  perimeter 
of  the  sector. 

707.  Compute  the  area  of  a  circle  in  which  the  chord,  3  feet 
long,  subtends  an  arc  of  120°. 

708.  Compute  the  area  of  a  segment  whose  arc  is  300°,  the 
radius  of  the  circle  being  50  cm. 

709.  The  areas  of  two  concentric  circles  are  as  5  to  8.  The 
area  of  that  part  of  the  ring  which  is  contained  between  two 
radii  making  the  angle  45°  is  300  square  feet.  Compute  the 
radii  of  the  two  circles. 

710.  What  is  the  altitude  of  a  rectangle  equivalent  to  a 
sector  whose  radius  is  15  m.,  if  the  base  of  the  rectangle  is 
equal  to  the  arc  of  the  sector? 

711.  Compute  the  radius  of  a  circle,  if  its  area  is  doubled  by- 
increasing  its  radius  one  foot. 

712.  The  radius  of  a  circle  is  400  cm.  Through  a  point 
exterior  to  the  circle  two  tangents  are  drawn,  making  an  angle 
of  60°.  Compute  the  area  of  the  figure  bounded  by  the  tangents 
and  the  intercepted  arc. 

713.  Three  equal  circles  are  drawn  tangent  to  each  other, 
with  a  radius  of  12  feet.  Compute  the  area  contained  between 
the  circles. 

714.  Upon  each  side  of  a  square,  as  a  diameter,  a  semi- 
circumference  is  described  within  the  square.  If  the  side  of 
the  square  is  10  meters,  compute  the  sum  of  the  areas  of  the 
four  leaves  in  ares. 

715.  In  a  circle  whose  radius  is  100  feet  two  parallel  chords 
are  drawn  on  the  same  side  of  the  center,  one  equal  to  the  side 


Problems  of  Computation.  125 

of  a  regular  hexagon,  and  the  other  to  the  side  of  a  regular 
trigon,  both  inscribed  in  the  given  circle.  Compute  the  area  of 
the  circle  comprised  between  the  two  parallel  chords. 

716.  A  quarter-mile  running  track  consists  of  two  parallel 
straight  portions  joined  together  at  the  ends  by  semicircum- 
f  erences.  The  extreme  length  of  the  plot  inclosed  by  the  track 
is  180  yards.  Compute  the  cost  of  sodding  this  plot  at  25  cents 
per  square  yard. 

717.  The  perimeter  of  a  certain  church  window  is  made  up 
of  three  equal  semicircumferences,  the  centers  of  which  form 
the  vertices  of  an  equilateral  triangle  whose  sides  are  1  m.  long. 
Compute  the  area  and  perimeter  of  the  window. 

718.  Two  flower  beds  have  equal  perimeters.  One  of  the 
beds  is  circular,  and  the  other  has  the  form  of  a  regular  hexa- 
gon. The  circular  bed  is  closely  surrounded  by  a  walk  7  feet 
wide.  The  area  of  the  walk  is  to  that  of  the  bed  as  7  to  9. 
Compute  the  diameter  of  the  circular  bed  and  the  area  of  the 
hexagonal  bed. 

719.  A  crescent-shaped  region  is  bounded  by  a  semicircum- 
ference  of  radius  a,  and  another  circular  arc,  the  center  of  the 
circumference  of  which  lies  on  the  semicircumference  produced. 
Compute  the  area  and  perimeter  of  the  crescent  region. 

720.  Six  coins,  each  of  radius  a,  are  placed  close  together  on 
a  table  so  that  their  centers  are  at  the  vertices  of  a  regular 
hexagon.  Compute  the  area  and  perimeter  of  the  inclosed 
figure. 

721.  Every  cross-section  of  the  train  house  of  a  railway  sta- 
tion has  the  form  of  a  pointed  arch  made  of  two  circular  arcs, 
the  centers  of  which  are  on  the  ground.  The  radius  of  each 
arc  is  equal  to  the  width  of  the  building,  210  feet.  Compute 
the  distance  across  the  building  measured  over  the  roof,  and 
show  that  the  area  of  the  cross-section  is  3675  (4  tt  —  3  V3) 
square  feet. 


126  Plane  Geometry. 


PROBLEMS    OP    CONSTRUCTION. 

722.  In  the  demonstration  of  the  foregoing  theorems  it  has 
been  assumed  that  certain  constructions  were  possible ;  i.e.  that 
perpendiculars  and  parallels  could  be  drawn,  that  lines  and 
angles  could  be  bisected,  etc.  It  is  now  proposed  to  show  that 
those  and  many  other  problems  can  be  performed,  so  our 
previous  demonstrations  are  not  vitiated  that  are  in  any  way 
dependent  upon  such  constructions. 

The  solution  of  a  geometrical  problem  of  construction  in- 
volves in  general  three  steps,  viz. : 

i.    The  construction  proper,  by  the  use  of  the  compass  and 

ruler. 
ii.   Demonstration  to  prove  the  correctness  of  the  construction, 
ill.   Discussion  of  its  limitations  and  applications,  including 

the  number  of  possible  constructions. 
If  numerical  or  algebraical  results  are  also  required,  then 
there  is,  of  necessity, 

iv.  Computation,  by  which  numerical  values  are  ascertained, 
involving  also  the  use  of  general  symbols  in  obtaining 
algebraic  formulae. 

Each  pupil  should  be  provided  with  a  good  pair  of  compasses, 
for  the  use  of  either  pen  and  ink,  or  pencil,  a  good  ruler  with 
straight  edges,  besides  one  hard  and  one  soft  pencil.  The 
graphite  of  the  pencil  should  be  sharpened  flat,  so  that  a  fine 
line  can  be  made. 

PROBLEMS. 

723.  It  is  required  to  find  a  point  which  is  a  given  distance  from  a 
given  point. 

Post.     Let  C  be  the  given  point,  and  AB  the  given  distance. 

We  are  required  to  find  a  point  which  shall  be  at  the  distance 
AB  from  (7. 

Cons.  First,  with  the  ruler,  from  C  draw  an  indefinite 
straight  line,  as  CD,  in  any  direction.     Then  with  the  com- 


Problems  of  Construction.  127 

passes,  using  (7  as  a  center  and  AB  as  a  radius,  draw  an  arc 
cutting  the  line  CDj  as  at  H. 


A 


If 
Then  H  is  the  required  point ;  for, 

Dem.  If,  in  applying  the  compasses  to  AB  21.  circle  had  been 
constructed,  and  the  circle  HKN  also  completed,  then  the 
circles  would  have  equal  radii. 

.-.   His,  the  same  distance  from  C that  A  is  from  B. 

Discussion.  Since,  from  the  definition  of  a  circle,  all  points 
in  the  circumference  are  equidistant  from  the  center,  it  follows 
that  every  point  in  the  circumference  HKN  is  the  same  distance 
from  C  that  A  is  from  B.  Consequently  any  point  in  that  cir- 
cumference answers  the  conditions  of  the  problem.  q.e.f. 

724.  Def.  Whenever  a  line  is  found  such  that  any  point  in 
it  selected  at  random  will  fulfill  certain  specified  conditions, 
or  such  that  all  points  in  it  have  a  common  property,  that  line 
is  called  the  locus  of  that  point  or  points.  Hence  we  may  say  in 
the  above  case,  that  the  circumference  HKN  is  the  locus  of  the 
point  which  is  the  distance  AB  from  point  C,  or,  as  some  prefer 
to  put  it,  it  is  the  locus  of  all  points  which  are  at  the  distance 
AB  from  point  C. 

725.  It  is  reqtdred  to  find  the  point,  having  given  its  distances  from 
two  given  points. 

A B 

G  D 

M : N 

Post.  Let  the  two  given  distances  be  AB  ai;d  MN,  and  the 
two  given  points  C  and  D. 


128  Plane  Geometry. 

We  are  required  to  find  a  point  which,  is  AB  distant  from  (7, 
and  MN  distant  from  D,  or  vice  versa. 

Sag.  Find  locus  of  the  point  which  is  AB  distant  from  point 
C,  and  locus  of  the  point  which  is  ifcO/"  distant  from  point  D,  and 
vice  versa. 

Dis.  How  many  points,  then,  satisfy  the  condition  of  the 
problem  ? 

Determine  the  result  if 
i.   the  distance  between  C  and  D  had  been  greater  than  the 

sum  of  AB  and  MJSf. 
ii.   If  it  had  been  equal  to  the  sum  of  AB  and  MJSf. 
iii.   If  it  had  been  equal  to  the  difference  of  AB  and  MN. 
iv.   If  it  had  been  less  than  the  difference  of  AB  and  MN. 
(See  Theorem  322.) 

726.  It  is  required  to  find  the  point  which  is  equidistant  from  two 
given  points. 

Post.     Let  A  and  B  be  the  two  given  points. 
We  are  to  find  a  point  which  is  the  same  distance  from  A  as 
from  B. 

It  is  evident  that  whether  there  be  more  or  not,  there  must 

at  least  be  one  midway  between 

■^  A  and  B ;  so  join  AB. 

\  j  y  With  A  and  B  as  centers,  con- 

•^X  struct,  with  the  same  radius,  two 

/  i  ^  circles  which  shall  intersect.   How 

I  can  you  tell  whether  or  not  they 

5  will  intersect  ? 

4 — n       Connect   the    points    of   inter- 

j  section,  as  D  and  H.     Then  DH 

\  sustains  what  relation  to  the  two 

j    ^  circles? 

\\/L  The    line    AB    sustains    what 

y  i  '^%,  relation  to  the  two  circles  ? 

;  Then  what  relation,  as  regards 

^l  position,  between  AB  and  DII? 


Problems  of  Construction.  129 

How  is  the  point  D  situated  with  reference  to  points  A  and  5? 

How  is  the  point  H  situated  with  reference  to  the  same 
points  ? 

Then  what  must  be  the  relation  of  AC  and  GB  as  regards 
magnitude  ? 

Suppose,  now,  DH  be  indefinitely  extended,  and  any  point 
in  it  selected  at  random.  How  will  this  point  be  situated  with 
reference  to  the  points  A  and  B  ?     Why  ? 

What  name,  then,  shall  we  give  to  the  line  JV7i  ?  Why  ?  q.e.f. 

727.  It  is  required  to  construct  the  perpendicular  bisector  of  a 
given  Hne. 

Sug.     Employ  method  similar  to  the  previous  one. 

728.  It  is  required  to  construct  a  perpendicular  to  a  given  straight 
line  which  shall  pass  through  a  given  point  in  that  hne. 

A  D  C  S  K 

Post.     Let  AK  be  the  given  line,  and  C  the  given  point. 

We  are  required  to  construct  a  perpendicular  to  AK  passing 
through  point  C. 

Sug.  Lay  off  equal  distances  each  side  of  C,  as  CD  and  CB ; 
then  use  Problem  727. 

729.  It  is  required  to  construct  a  perpendicular  to  a  line  at  one 
extremity. 

Sug.  Extend  the  line ;  then  use  previous  problem.  In  case 
the  extension  should  not  be  possible  , 

or  convenient,  use  the  following :  2^"*n  { jr 

Select  any  point  at  random,  as  C,  /Yn 

making  sure  that  it  lies  between  A  /     \    \ 

and  B.     Then,  with  CB  as  a  radius,  c/  \     \ 

construct   the   circle    or   arc   KBN.  /''  \     I 

Through  D,  point  where  this  circle        js:^  /  \  / 

intersects  the  given  line,  and  (7,  draw  A.     i/x^    ~7B 

the  diameter  BH.     Join  HB.  ^"^ '' 

Then  KB  is  the  perpendicular  required.     Why  ?  q.e.f. 


130  Plane  Geometry. 

730.   It  is  required  to  construct  a  perpendicular  to  a  given  line  from 
a  given  point  without  tlie  line. 

G  ' 


^ / 


Post.    Let  AB  be  the  given  line  and  C  the  given  point. 
8ug.     With  (7  as  a  center,  construct  a  circle  which  shall 
intersect  AB,  as  KDHN.     Use  Problem  727. 

731.  It  is  required  to  bisect  a  given  arc. 

Sug.  Connect  the  extremities  of  the  given  arc;  then  use 
Problem  727,  and  for  proof  consult  Theorems  310  and  299. 

732.  It  is  required  to  bisect  a  given  angle. 
Sug.     Consult  Theorem  95. 

733.  At  a  given  point  in  a  given  line,  it  is  required  to  construct  an 
angle  equal  to  a  given  angle,  the  given  line  forming  one  side  of  the 
angle. 

Sug.     Consult  293  and  292. 

734.  It  is  required  to  draw  through  a  given  point  a  line  parallel  to 
a  given  line. 

Sug.     Consult  115  or  108,  and  Problem  733. 

735.  Two  angles  of  a  triangle  being  given,  it  is  required  to  construct 
the  third  angle. 

Sug.     Consult  Theorems  126  and  72,  and  Problem  733. 

736.  Having  given  two  sides  of  a  triangle  and  their  included  angle, 
it  is  required  to  construct  the  triangle. 

737.  Having  given  two  angles  of  a  triangle  and  the  side  joining 
their  vertices,  it  is  required  to  construct  the  triangle. 

When  is  this  problem  impossible  ? 


Problems  of  Construction.  131 

738.  Having  given  three  sides  of  a  triangle,  it  is  required  to  con- 
struct the  triangle. 

When  is  this  problem  impossible  ? 

739.  It  is  required  to  construct  the  locus  of  the  point  which 
is  a  given  distance  from  a  given  straight  line. 

740.  It  is  required  to  construct  a  locus  of  the  point  which  is 
a  given  distance  from  a  given  circumference. 

741.  It  is  required  to  construct  the  locus  of  a  point  which  is 
equally  distant  from  two  given  parallel  lines. 

742.  It  is  required  to  construct  the  locus  of  the  point  which 
is  equally  distant  from  two  non-parallel  lines  in  the  same 
plane. 

743.  It  is  required  to  construct  the  locus  of  the  point  which 
is  equally  distant  from  the  circumferences  of  two  equal  circles. 

744.  Having  given  the  hypothenuse  of  a  right  triangle,  it  is 
required  to  construct  the  locus  of  the  vertex  of  the  light  angle. 

Sug.     Consult  Theorem  394. 

745.  It  is  required  to  find  in  a  given  line,  a  point  which  is 
equally  distant  from  two  given  points. 

Sug.     Consult  Theorem  100. 

746.  It  is  required  to  find  a  point  which  is  equidistant  from  three 
given  points. 

Sug.     Join  two  pairs  of  points,  then  consult  Theorem  100. 

747.  Through  a  given  point  without  a  given  line,  it  is  required  to 
draw  a  fine  which  shall  make  an  angle  with  the  given  line  equal  to  a 
given  angle. 

748.  It  is  required  to  construct  the  triangle,  having  given  the  base, 
the  vertical  angle,  and  one  of  the  other  angles. 

749.  It  is  required  to  construct  the  triangle,  having  given  two  sides 
and  an  angle  opposite  one  of  them. 

How  many  possible  constructions  ? 


132  Plane  Geometry. 

750.  It  is  required  to  find  the  center  of  a  circle,  having  the  circum- 
ference or  an  arc  given. 

751.  It  is  required  to  construct  the  circumference,  having  given 
three  points  in  it. 

752.  It  is  required  to  construct  a  circumference  which  shall  pass 
through  the  vertices  of  a  given  triangle, 

753.  It  is  required  to  find  the  locus  of  the  center  of  the  circumfer- 
ence which  shall  pass  through  two  given  points. 

754.  With  a  given  radius,  it  is  required  to  construct  the  circle 
which  shall  pass  through  two  given  points. 

755.  It  is  required  to  construct  the  isosceles  triangle,  having  given 
the  base  and  the  verticle  angle. 

756.  It  is  required  to  construct  a  circumference  which  shall 
be  a  given  distance  from  three  given  points. 

757.  It  is  required  to  construct  a  circle  which  shall  have  its 
center  in  a  given  straight  line  and  circumference  passing  through 
two  given  points. 

758.  It  is  required  to  find  a  point  which  shall  be  equidistant 
from  two  given  points,  and  at  a  given  distance  from  a  third 
given  point. 

759.  It  is  required  to  construct  the  equilateral  triangle,  having 
given  one  side. 

760.  It  is  required  to  trisect  a  right  angle. 

761.  It  is  required  to  find  a  point  which  shall  be  equally- 
distant  from  two  given  points,  and  also  equally  distant  from 
two  given  parallel  lines. 

762.  It  is  required  to  find  a  point  which  shall  be  equally 
distant  from  two  given  points,  and  also  equally  distant  from 
two  given  non-parallel  lines  in  the  same  plane. 

763.  It  is  required  to  find  a  point  which  shall  be  equally 
distant  from  two  given  parallel  lines,  and  also  equally  distant 
from  two  non-parallel  lines  in  the  same  plane. 


Problems  of  Construction.  133 


764. 

It  is  required  to  construct 

i. 

an  angle  of  45° ; 

vi. 

an  angle  of  75° ; 

ii. 

an  angle  of  60°  ; 

vii. 

an  angle  of  22°  30' ; 

iii. 

an  angle  of  30° ; 

viii. 

an  angle  of  52°  30'; 

iv. 

an  angle  of  15° ; 

ix. 

an  angle  of  135° ; 

V. 

an  angle  of  105° ; 

X. 

an  angle  of  165°. 

765.  It  is  required  to  find  a  point  in  one  side  of  a  triangle 
which  shall  be  equally  distant  from  the  other  two  sides. 

766.  It  is  required  to  find  a  point  which  shall  be  equally- 
distant  from  two  non-parallel  lines  in  the  same  plane,  and  at  a 
given  distance  from  a  given  point. 

767.  It  is  required  to  construct  the  right  triangle,  having  given  the 
two  legs. 

768.  It  is  required  to  construct  the  right  triangle,  having  given  the 
hypothenuse  and  one  acute  angle. 

769.  It  is  required  to  construct  the  right  triangle,  having  given  one 
leg  and  adjacent  acute  angle, 

770.  It  is  required  to  construct  the  right  triangle,  having  given  one 
leg  and  the  acute  angle  opposite. 

771.  It  is  required  to  construct  the  right  triangle,  having  given  the 
hypothenuse  and  one  leg. 

772.  It  is  required  to  construct  a  tangent  to  a  given  circle  passing 
through  a  given  point. 

Case    I.   When  the  given  point  is  the  point  of  contact. 

Case  II.   When  the  given  point  is  outside  the  circle. 

Sug.  Connect  the  center  of  the  given  circle  with  the  given 
exterior  point.  On  this  line  as  a  diameter  construct  a  circle, 
and  join  the  points  of  its  intersection  of  the  given  circle  with 
extremities  of  the  diameter. 

773.  It  is  required  to  construct  the  parallelogram,  having  given 
two  sides  and  their  included  angle. 


134 


Plane  Geometry. 


774.  It  is  required  to  construct  a  circle  within  a  given  triangle,  so 
that  the  sides  of  the  triangle  shall  be  tangents  of  the  circle,  i.e.  to  in- 
scribe a  circle. 

Sug.   Consult  Theorems  146  and  149. 

775.  It  is  required  to  construct  the  circumference  of  a  circle,  having 
given  a  chord  and  angle  made  by  the  chord  and  a  tangent  to  that  circle. 


N'   K 


Let  AB  be  the  given  chord,  and  C  the  angle  made  by  the 
chord  and  tangent.  Then  one  extremity  of  the  chord  must  be 
the  point  of  contact. 

Take  DH  =  AB.  Construct  angle  DHN  equal  to  angle  C, 
and  draw  PHI.  to  NH. 

Then  H  is  the  point  of  contact,  and  NH  the  tangent. 

Find  the  locus  of  the  center  of  the  circumference  passing 
through  D  and  H.     Problem  753. 

Hence  T  must  be  the  center  of  the  circle  required. 

If  from  any  point  in  the  arc  DOH  (call  the  point  Q)  lines  be 
drawn  to  D  and  H,  how  will  the  angle  Q  compare  in  magnitude 
with  the  angle  DHN? 

What  is  formed  by  the  lines  DH,  DQ,  and  HQ? 

If  we  call  DH  the  base,  what  would  you  call  the  angle  Q  ? 


Problems  of  Construction.  135 

What  the  point  Q? 

What  might  the  arc  DQHhe  named,  then?  q.e.f. 

776.  Having  given  the  base  and  vertical  angle  of  a  triangle, 

it  is  required  to  construct  the  locus  of  its  vertex. 
Sug.     Consult  the  previous  problem. 

777.  It  is  required  to  construct  the  triangle,  having  given  the  base, 
the  vertical  angle,  and  the  altitude. 

Sug.     Use  previous  problem  or  775. 

778.  It  is  required  to  construct  the  triangle,  having  given  the  base, 
the  vertical  angle,  and  the  median, 

OPTIONAL  PROBLEMS  FOR  ADVANCED  WORK. 

779.  It  is  required  to  construct  the  triangle,  having  given 
the  base,  vertical  angle,  and  perpendicular  from  one  extremity 
of  the  base  to  the  opposite  side. 

780.  It  is  required  to  construct  the  isosceles  triangle,  having 
given  the  altitude  and  one  of  the  equal  angles. 

781.  It  is  required  to  construct  the  chord  in  a  given  circle, 
having  given  the  middle  point  of  the  chord. 

782.  It  is  required  to  construct  a  circle  whose  circumference 
shall  pass  through  the  vertices  of  a  given  rectangle. 

783.  It  is  required  to  construct  a  tangent  to  a  given  circle 
which  shall  be  parallel  to  a  given  straight  line. 

784.  It  is  required  to  construct  a  tangent  to  a  given  circle 
which  shall  be  perpendicular  to  a  given  straight  line. 

785.  It  is  required  to  construct  a  tangent  to  a  given  circle 
which  shall  make  a  given  angle  with  a  given  straight  line. 

786.  It  is  required  to  construct  the  isosceles  triangle,  having 
given  the  vertical  angle,  and  a  point  in  the  base,  in  position. 

787.  It  is  required  to  construct  the  triangle,  having  given 
the  altitude,  the  base,  and  an  adjacent  angle. 


136  Plane  Geometry. 

788.  It  is  required  to  construct  the  triangle,  having  given 
the  altitude,  the  base,  and  an  adjacent  side. 

789.  It  is  required  to  construct  a  rhombus,  having  given  its 
base  and  altitude. 

790.  It  is  required  to  construct  the  triangle,  having  given 
the  altitude  and  the  sides  which  include  the  vertical  angle. 

791.  It  is  required  to  construct  the  triangle,  having  given 
the  altitude  and  angles  adjacent  to  the  base. 

792.  It  is  required  to  construct  an  isosceles  triangle  which 
shall  have  its  vertical  angle  twice  the  sum  of  its  other  two 
angles. 

793.  It  is  required  to  construct  the  square,  having  given  its 
diagonal. 

794.  It  is  required  to  construct  the  right  triangle,  having 
given  the  hypothenuse  and  perpendicular  from  the  vertex  of 
the  right  angle  to  the  hypothenuse. 

795.  ■  It  is  required  to  construct  the  locus  of  the  center  of  the 
circle  of  given  radius  tangent  to  a  given  straight  line. 

796.  It  is  required  to  construct  the  circle  of  given  radius 
which  shall  be  tangent  to  two  given  non-parallel  lines. 

797.  It  is  required  to  construct  the  circle  of  given  radius 
which  shall  be  tangent  to  a  given  straight  line,  and  whose 
center  shall  be  in  a  given  line  not  parallel  to  the  former. 

798.  It  is  required  to  construct  the  circle  which  shall  be 
tangent  to  a  given  line,  and  whose  circumference  shall  pass 
through  a  given  point. 

799.  It  is  required  to  find  the  locus  of  the  center  of  the 
circle  of  a  given  radius  which  shall  be  tangent  externally  to  a 
given  circle. 


Problems  of  Construction.  137 

800.  It  is  required  to  construct  the  locus  of  the  center  of  the 
circle  of  given  radius  which  shall  be  tangent  internally  to  a 
given  circle. 

801.  It  is  required  to  construct  a  circle  with  a  given  radius 
which  shall  be  tangent  to  a  given  line  and  a  given  circle. 

802.  It  is  required  to  construct  a  circle  with  a  given  radius 
which  shall  be  tangent  to  two  given  circles. 

803.  It  is  required  to  construct  a  circle  which  shall  cut 
three  equal  chords  of  given  length  from  three  given  non- 
parallel  lines. 

804.  It  is  required  to  construct  in  a  given  circle  a  chord  of 
given  length  passing  through  a  given  point. 

805.  It  is  required  to  construct  in  a  given  circle  a  chord  of 
given  length  and  parallel  to  a  given  straight  line. 

806.  It  is  required  to  construct  a  line  of  given  length  passing 
through  a  given  point  between  two  given  parallel  lines. 

807.  It  is  required  to  construct  a  line  of  given  length 
between  two  given  non-parallel  lines,  and  which  shall  be 
parallel  to  a  given  line. 

808.  It  is  required  to  construct  the  right  triangle,  having 
given  the  hypothenuse  and  radius  of  the  inscribed  circle. 

809.  It  is  required  to  construct  the  right  triangle,  having 
given  the  radius  of  the  inscribed  circle  and  one  acute  angle. 

810.  It  is  required  to  construct  the  triangle,  having  given 
in  position  the  middle  points  of  its  sides. 

811.  It  is  required  to  construct  the  triangle,  having  given 
the  base,  vertical  angle,  and  radius  of  the  circumscribed  circle. 

812.  It  is  required  to  construct  the  isosceles  triangle,  having 
given  the  base  and  radius  of  the  inscribed  circle. 


138  Plane  Geometry. 

813.  It  is  required  to  construct  a  straight  line  which  shall 
pass  through  a  given  point  and  make  equal  angles  with  two 
given  lines. 

814.  It  is  required  to  find  a  point  in  a  given  secant  to  a 
given  circle  such  that  the  tangent  to  the  circle  from  that  point 
shall  be  of  given  length. 

815.  It  is  required  to  construct  the  right  triangle,  having 
given  one  leg  and  radius  of  the  inscribed  circle. 

816.  It  is  required  to  construct  the  right  triangle,  having 
given  the  median  and  altitude  from  the  vertex  of  the  right 
angle  to  the  hypothenuse. 

817.  It  is  required  to  find  the  locus  of  the  center  of  the 
chord  which  passes  through  a  given  point  in  a  given  circle. 

818.  It  is  required  to  construct  the  triangle,  having  given  the 
base,  vertical  angle,  and  sum  of  the  other  two  sides. 

819.  It  is  required  to  construct  a  circle  which  shall  be  tan- 
gent to  two  given  lines  and  at  given  point  of  contact  in  one. 

820.  It  is  required  to  inscribe  a  circle  in  a  given  sector. 

821.  It  is  required  to  construct  a  common  tangent  to  two 
given  circles. 

Sug.  Make  five  cases  according  to  the  relative  position  of 
the  circles.     (See  Theorem  322.) 

822.  It  is  required  to  inscribe  a  square  in  a  given  rhombus. 

823.  Having  given  two  intersecting  circles,  it  is  required  to 
draw  a  line  through  one  of  the  points  of  intersection,  so  that 
the  two  intercepted  chords  shall  be  equal. 

Sug.  Join  centers.  Draw  from  point  of  intersection  to 
middle  of  that  line. 

From  centers  draw  radii  parallel  to  latter  line.  Through 
point  of  intersection  draw  line  perpendicular  to  these  radii. 


Problems  involving  Ratio.  139 

824.  It  is  required  to  construct  an  equilateral  triangle  hav- 
ing its  vertices  in  three  given  parallel  lines. 

825.  It  is  required  to  construct  a  tangent  to  a  given  circle 
with  two  given  parallel  secants,  so  that  the  point  of  contact 
shall  bisect  the  part  between  the  secants. 

826.  It  is  required  to  construct  three  equal  circles  which 
shall  be  tangent  to  each  other  and  also  to  a  given  circle 
externally. 

827.  It  is  required  to  construct  three  equal  circles  which 
shall  be  tangent  to  each  other  and  also  to  a  given  circle 
internally. 

828.  It  is  required  to  construct  three  equal  circles  which 
shall  be  tangent  to  each  other  and  also  to  the  sides  of  an 
equilateral  triangle. 

829.  It  is  required  to  construct  a  semicircle  having  its 
diameter  in  one  of  the  sides  of  a  given  triangle  and  tangent 
to  the  other  two  sides. 

830.  It  is  required  to  construct  a  triangle,  having  given  the 
radius  of  the  inscribed  circle  and  two  sides. 

831.  It  is  required  to  find  a  point  in  a  given  line  such  that 
lines  to  that  point  from  two  given  points  without  the  line 
make  equal  angles  with  the  line. 

832.  It  is  required  to  construct  the  triangle,  having  given 
the  perimeter,  altitude,  and  vertical  angle. 

PROBLEMS  OF   CONSTRUCTION,   INVOLVING  RATIO  AND 
PROPORTION. 

It  is  required, 

833.  To  divide  a  given  Hne  into  any  number  of  equal  parts. 

Sug.  From  one  extremity  of  the  given  line  construct  a  line 
of  indefinite  length,  making  any  convenient  angle  with  the  given 
line.     Then,  with  any  convenient  unit  of  length  assumed  as  a 


140  Plane  Geometry. 

unit  of  measure,  beginning  at  the  vertex,  lay  off  on  the  indefi- 
nite line  this  unit  as  many  times  as  it  is  required  to  divide  the 
given  line  into  parts.     Then  consult  422. 

834.  To  divide  a  given  line  into  parts  proportional  to  any  number 
of  given  lines. 

Sug.  Draw  a  line  as  in  833  and  set  off  on  this  line  segments 
equal  to  the  given  lines.     Then  consult  427. 

835.  To  construct  a  line  that  shall  be  a  fourth  proportional  to  three 
given  lines. 

Sug.     Consult  Theorem  423. 

836.  To  construct  a  line  that  shall  be  a  third  proportional  to  two 
given  lines. 

Sug.  What  must  one  of  the  given  lines  be  if  those  two  and 
one  other  are  to  form  a  proportion?  Consider  that  in  the 
construction. 

837.  To  construct  a  mean  proportional  between  two  given  lines. 
Sug.     Consult  Theorem  449. 

838.  Given  a  polygon  and  an  homologous  side  of  another  similar 
polygon,  to  construct  the  latter. 

Sug.     Consult  Theorem  477. 

839.  To  inscribe  in  a  given  circle  a  triangle  similar  to  a 
given  triangle. 

840.  To  circumscribe  about  a  given  circle  a  triangle  similar 
to  a  given  triangle. 

841.  To  construct  a  square  which  shall  be  equivalent  to  the  sum  of 
two  given  squares. 

Sug.     Consult  Theorem  525. 

842.  To  construct  a  square  which  shall  be  equivalent  to  the  sum  of 
three  or  more  given  squares. 

Sug.  Construct  a  square  equivalent  to  two  of  the  given 
squares,  then  one  equivalent  to  that  and  one  other  of  the  given 
squares,  and  so  on. 


Problems  involving  Ratio.  141 

843.  To  construct  a  square  which  shall  be  equivalent  to  the  differ- 
ence of  two  given  squares. 

844.  To  construct  a  square  equivalent  to  a  given  rectangle. 

Sug.     If  X  and  y  are  the  base  and  altitude  of  a  rectangle, 
and  n  one  side  of  an  equivalent  square,  then 

xy  =  n^. 

Whence  x:n::n:y     (see  Theorem  357) 

or  71  is  a  mean  proportional  between  x  and  y. 
Hence  consult  Problem  837. 

845.  To  construct  a  square  which  shall  be  equivalent  to  a  given 
parallelogram. 

846.  To  construct  a  square  which  shall  be  equivalent  to  a  given 
triangle. 

847.  To  construct  a  triangle  which  shall  be  equivalent  to  a  given 
polygon  of  more  than  three  sides. 

Post.     Let  ABCDH  be  a  polygon  of  n  sides. 

Extend  one  of  the  sides  _^ 

as   ABy   and   construct   the  /"V. 

diagonal  DB.     Through  ver-  /       V*N. 

tex  G  draw  CK  parallel  to        /  \    vs. 

i)5  and  join  i)/t:  ^\  \    XV 

Considering  DB  the  com-  \  \       ^y^x 

mon   base  of    the   two  tri-  \     ^y       "\b: 

angles  DOB  and  DKB,  what  ^  ^  * 

is  the  relation  between  the  areas  of  those  two  triangles? 
Why? 

How  does  the  area  of  the  polygon  DKAH  then  compare 
with  that  of  DOB  AH?    Why  ? 

How  many  sides  has  the  former  as  compared  with  the  latter  ? 

Proceed  in  the  same  way  with  the  polygon  DKAH. 

848.  To  construct  a  square  equivalent  to  any  given  polygon, 
Sug.     Use  Problems  847  and  846. 


142  Plane  Geometry. 

849.  To  construct  a  square  equivalent  to  the  sum  of  any  number  of 
given  polygons. 

850.  To  construct  a  rectangle  wMch  shall  be  equivalent  to  a  given 
square,  and  the  sum  of  whose  base  and  altitude  shall  be  equal  to  a 
given  line. 

Sug.  Upon  the  given  line  as  a  diameter  construct  a  circle. 
Construct  a  line  parallel  to  the  diameter,  distant  from  it  one 
side  of  the  given  square.     Then  consult  Theorem  449. 

851.  To  construct  a  rectangle  which  shall  be  equivalent  to  a  given 
square,  and  the  difference  of  whose  base  and  altitude  shall  be  equal  to 
a  given  line. 

Sug.  Proceed  as  in  850,  then  at  extremity  of  the  diameter 
construct  a  tangent  equal  to  one  side  of  a  given  square,  and 
from  other  extremity  of  this  tangent  construct  a  secant  through 
center  of  circle.     (Consult  Theorem  452.) 

852.  To  construct  a  right  triangle  which  shall  be  equivalent 
to  a  given  triangle  and  its  hypothenuse  equal  to  a  given  line. 

853.  To  divide  a  given  line  into  extreme  and  mean  ratio.    (See  348.) 

N 


Jl S  j3^ 

Post.  Let  AB  be  the  given  line.  At  one  extremity  erect  a 
perpendicular  CB  equal  to  one  half  AB.  With  O  as  a  center 
and  CB  as  a  radius  construct  the  circle  DBN.  Construct  the 
secant  AK  passing  through  the  center  C.     With  ^  as  a  center 


Problems  involving  Ratio.  143 

and  radius  ADj  construct  the  arc  DR.     Then  the  line  AB  will 
be  divided  in  extreme  and  mean  ratio  at  H. 

Dem. 


Since            CB-. 

=  f ,    DK. 

=  AB. 

ADiAB: 

:AB:AK 

Why? 

AD:AB-AD: 

lABiAK- 

■AB 

Why? 

AH:AB-AH: 

:AB:AK- 

-DK 

Why? 

AHiHB: 

lABiAD 

Why? 

AHiHB: 

:AB:AH 

Why? 

HB'.AH: 

:AH:AB 

Why? 

or 
Hence  the  line  AB  is  divided  in  extreme  and  mean  ratio. 

Q.E.F. 

854.  To  inscribe  a  regular  decagon  in  a  given  circle. 

Post.    Let  ABH  be  the  given  circle,  and  AC  one  of  its  radii. 

Consult  Problem  853. 

Cons.  With  J.  as  a  center  and 
DC  as  a  radius,  construct  chord 
AB.    Join  BC  and  BD. 

Dem. 
AC:AB::AB:AD    Why? 

Hence  the  two  A  ABD  and 
ABC  are  how  related?  (See 
Theorem  441.) 

What  kind  of  a  triangle  is  ACB  ? 

What  kind  of  a  triangle  must  ABD  be,  then  ? 

What  relation,  then,  between  AB  and  DB  ?  What  between 
JBDandDO? 

What  relation  between  the  A  CAB  and  CBA  ?  Between  the 
A  DAB  and  BDA  ?  Between  the  A  BDA  and  CBA  ?  Why  ? 
Between  the  A  DBC  and  DCB  ? 

What  relation  does  the  Z  BDA  bear  to  the  sum  of  DCB  and 
DBC?    Why? 


144  Plane  Geometry. 

What  relation  does  Z  DAB  bear  to  Z  (7,  then  ? 

Hence,  what  relation  does  Z  ABC  bear  to  Z  (7? 

Compare  now  Z  DAB  4-  Z  ^jB(7  with  the  Z  (7,  and  finally 
compare  Z  i>^5  +  Z  ABC+  Z  O  with  the  Z  (7. 

If  the  pupil  has  answered  the  above  questions  correctly,  he 
will  have  now  the  equation 

Z  i)^^  + Z  ^5(7  + Z  (7=  5  Z  C. 

What  is  the  value  of  the  first  member  of  the  above  equa- 
tion ?    Why  ? 

Then  5ZC=2Tt.A 

or  10ZO=4rt.  Zs 

Whence  Z  0  =  z^-  of  4  rt.  ^. 

Hence  the  arc  AB  is  what  part  of  the  circumference  ? 

.-.  AB  is  the  side  of  a  regular  inscribed  decagon.  q.e.f. 

855.  To  inscribe  a  square  in  a  given  circle, 
JS2ig.     Consult  Theorem  639. 

856.  To  inscribe  a  regular  hexagon  in  a  given  circle. 
JSug.   Consult  Theorem  471. 

857.  To  inscribe  a  regular  pente decagon  in  a  given  circle. 

Sug.  Find  the  difference  between  the  central  angle  of  the 
regular  decagon  and  that  of  a  regular  hexagon. 

858.  To  inscribe  in  a  given  circle 


i.  a  regular  trigon ; 

ii.  a  regular  pentagon ; 

ill.  a  regular  octagon ; 

iv.  a  regular  dodecagon  ; 


V.  a  regular  polygon  of 
sixteen  sides; 

vi.  a  regular  polygon  of 
twenty  sides. 


859.  To  circumscribe  about  a  given  circle  all  the  above-mentioned 
regular  polygons. 

860.  To  inscribe  in  a  given  circle  a  regular  polygon  similar  to  a 
given  regular  polygon. 


Problems  involving  Ratio. 


H5 


Sug.   Construct  a  central  angle  equal  to  that  of  the  given 
polygon,  etc. 

861.  To  circumscribe  a  circle  about  any  given  regular  polygon. 

862.  Upon  a  given  line  as  a  base,  to  construct  a  rectangle  equivalent 
to  a  given  rectangle. 

863.  To  construct  a  square  whose  ratio  to  a  given  square  shall  be 
the  same  as  that  of  two  given  lines. 


Post.  Let  Q  be  the  given  square,  and  I  and  p  the  two  given 
lines. 

"VYe  are  required  to  construct  a  square  (Q')  so  that 

Area  Q :  area  Q'  ::l:p. 

Cons.     Draw  AB  =  Z  +  p. 

On  this  as  a  diameter  construct  a  semicircle,  and  at  D  erect 
the  perpendicular  DK. 

Join  AK  and  BK.  Make  KN  equal  to  one  side  of  the  given 
square,  as  ST. 

Draw  NH  parallel  to  AB.  Then  KH  is  the  side  of  the 
required  square. 

Dem.  AK''  :  BK^  ::AD:DB  (?) 

or  AK^:BK^::l:p  (?) 

Again,  NK:  HK  :  :  AK:  BK  (?) 

Hence  NK"  :  HK'  :  :  AK"  :  BK"  (?) 


.-.  NK'.HK'.'.l'.p  (?) 

Hence  the  square  constructed  on  HK  as  a  side  is  the  re- 
quired square.  q.e.f. 


146 


Plane  Geometry. 


864.   To  construct  a  polygon  similar  to  a  given  polygon,  the  ratio 
of  whose  areas  sliall  be  that  of  two  given  lines. 


Post  Let  ABCDH  be  the  given  polygon  (P),  and  p  and  q 
the  two  given  lines. 

We  are  required  to  construct  a  polygon  {P)  similar  to  P,  so 

*^^^  Area  P:  area  P'::p:g. 

Cons.  Upon  any  side  of  the  polygon  P,  as  AB,  construct  a 
square. 

Then,  by  the  previous  problem,  find  the  side  of  a  square 
whose  ratio  to  that  of  the  square  on  AB  shall  be  that  of  the 
two  lines  p  and  q. 

Upon  this  line  construct  a  polygon  similar  to  polygon  P. 
This  will  be  the  polygon  required. 

Dem.    This  will  be  left  for  the  pupil. 

865.  To  construct  a  polygon  similar  to  one  of  two  given  polygons 
and  equivalent  to  the  other, 


Post    Let  P  and  Q  be  the  two  given  polygons.     We  are 
required  to  construct  a  polygon  similar  to  P  and  equivalent  to  Q. 


Problems  involving  Ratio.  147 

Cons.  Construct  squares  equivalent  to  each  of  the  polygons 
P  and  Q. 

Then  find  a  fourth  proportional  to  the  sides  of  these  squares 
and  any  side  of  the  polygon  P  selected  at  random,  as  AB. 
Upon  this  fourth  proportional  as  an  homologous  side  construct 
a  polygon  similar  to  P.  Then  this  polygon  will  be  similar  to 
P  and  equivalent  to  Q,  and  is  therefore  the  polygon  required. 

Dem.     This  is  also  left  for  the  pupil. 

866.  To  construct  upon  a  given  line,  as  one  side, 

i.  a  regular  trigon ;  v.  a  regular  pentagon ; 

ii.  a  regular  tetragon ;  vi.  a  regular  decagon ; 

iii.  a  regular  hexagon ;  vii.  a  regular  dodecagon ; 

iv.  a  regular  octagon ;  viii.  a  regular  pentedecagon. 

867.  To  construct  a  regular  hexagon,  given  one  of  its  shorter 
diaeronals. 


To  construct  a  regular   pentagon,   given   one   of   its 
diagonals. 

869.  To  construct  a  circle  equivalent  to  the  sum  of  two 
given  circles. 

870.  To  construct  a  circumference  equal  to  the  sum  of  two 
given  circumferences. 

871.  To  divide  a  given  circle  by  a  concentric  circumference 
into  two  equivalent  parts. 

MISCELLANEOUS   PROBLEMS   FOR  ADVANCE  WORK. 

I.   Triangles. 

It  is  required  to  construct  the  triangle,  having  given 

872.  Its  base,  vertical  angle^  and  difference  of  the  other  two 

sides. 

873.  Its  base,  vertical  angle,  and  a  square  which  is  equal  to 
the  sum  of  the  squares  upon  the  other  two  sides. 


148  Plane  Geometry. 

874.  Its  base,  vertical  angle,  and  a  square  "which  is  equiva- 
lent to  the  difference  of  the  squares  upon  the  other  two  sides. 

875.  Its  base,  vertical  angle,  and  sum  of  its  altitude  and  the 
two  remaining  sides. 

876.  Its  base,  vertical  angle,  and  the  sum  of  its  altitude  and 
difference  of  the  other  two  sides. 

877.  Its  base,  vertical  angle,  and  difference  between  its  alti- 
tude and  sum  of  its  other  two  sides. 

878.  Its  base,  vertical  angle,  and  difference  between  its  alti- 
tude and  difference  of  the  other  two  sides. 

879.  Its  base,  vertical  angle,  and  ratio  of  its  altitude  to  the 
sum  of  its  other  two  sides. 

880.  Its  base,  vertical  angle,  and  ratio  of  its  altitude  to  the 
difference  of  its  other  two  sides. 

881.  Its  base,  altitude,  and  sum  of  its  other  two  sides. 

882.  Its  base,  altitude,  and  difference  of  its  other  two  sides. 

883.  Its  base,  altitude,  and  ratio  of  the  other  two  sides. 

884.  Its  base,  altitude,  and  a  square  equivalent  to  the  rec- 
tangle of  the  other  two  sides. 

885.  Its  base,  altitude,  and  a  square  which  is  equivalent  to 
the  sum  of  the  squares  upon  the  other  two  sides. 

886.  Its  base,  altitude,  and  a  square  which  is  equivalent  to 
the  difference  of  the  squares  upon  the  other  two  sides. 

887.  Its  vertical  angle,  sum  of  base  and  altitude,  and  sum 
of  the  other  two  sides. 

888.  Its  vertical  angle,  sum  of  base  and  altitude,  and  differ- 
ence of  its  other  two  sides. 

889.  Its  vertical  angle,  sum  of  base  and  altitude,  and  ratio 
of  the  other  two  sides. 


Problems  in  Triangles.  149 

890.  Its  vertical  angle,  sum  of  base  and  altitude,  and  a  square 
equivalent  to  the  rectangle  of  its  other  two  sides. 

891.  Its  vertical  angle,  sum  of  base  and  altitude,  and  sum  of 
the  three  sides. 

892.  Its  vertical  angle,  sum  of  base  and  altitude,  and  differ- 
ence between  the  base  and  sum  of  the  other  two  sides. 

893.  Its  vertical  angle,  sum  of  base  and  altitude,  and  differ- 
ence between  the  base  and  difference  of  its  other  two  sides. 

894.  Its  vertical  angle,  sum  of  base  and  altitude,  and  the 
ratio  of  the  base  to  the  sum  of  the  other  two  sides. 

895.  Its  vertical  angle,  sum  of  base  and  altitude,  and  the 
ratio  of  the  base  to  the  difference  of  its  other  two  sides. 

896.  Its  vertical  angle,  altitude,  and  radius  of  the  circum- 
scribed circle. 

897.  Its  vertical  angle,  radius  of  the  inscribed  circle,  and 
perimeter. 

898.  Its  vertical  angle,  radius  of  the  inscribed  circle,  and 
ratio  of  the  sides  including  the  vertical  angle. 

899.  Its  vertical  angle,  radius  of  the  inscribed  circle,  and  a 
square  equivalent  to  the  rectangle  of  the  sum  of  the  two  sides 
including  the  vertical  angle  and  the  base. 

900.  Its  vertical  angle,  radius  of  the  inscribed  circle,  and  a 
square  whose  area  is  equal  to  the  difference  between  the  sum  of 
the  squares  of  the  sides  including  the  vertical  angle  and  the 
square  of  the  base. 

901.  Its  base,  median,  and  sum  of  the  other  two  sides. 

902.  Its  base,  median,  and  difference  of  the  other  two  sides. 

903.  Its  three  altitudes. 

904.  Its  three  medians. 

905.  Two  sides,  and  difference  of  the  angles  opposite  them. 


150  Plane  Geometry. 

906.  Its  vertical  angle,  difference  of  the  angles  at  the  base, 
and  difference  of  the  other  two  sides. 

907.  Difference  of  the  angles  at  the  base,  difference  of  the 
segments  of  the  base  made  by  the  altitude,  and  sum  of  the 
other  two  sides. 

908.  It  is  required  to  construct  the  equilateral  triangle, 
having  given  the  three  distances  from  the  vertices  to  a  common 
point. 

II.  Quadrilaterals. 

909.  It  is  required  to  construct  a  square,  having  given 
i.   The  sum  of  its  diagonal  and  side. 

ii.  The  difference  of  its  diagonal  and  side. 

910.  It  is  required  to  construct  a  rectangle,  having  given 
i.   The  sum  of  two  adjacent  sides  and  its  diagonal. 

ii.   The  difference  of  two  adjacent  sides  and  its  diagonal, 
iii.   One  side,  and  sum  of  diagonal  and  adjacent  side. 
iv.   One  side,  and  difference  of  diagonal  and  adjacent  side. 

911.  It  is  required  to  construct  the  rhombus,  having  given 
i.   Its  altitude  and  lesser  angle. 

ii.   Its  side  and  sum  of  its  diagonals. 

iii.   Its  side  and  difference  of  its  diagonals. 

iv.   Its  lesser  angle  and  sum  of  its  diagonals. 

V.   Its  lesser  angle  and  difference  between  its  longer  diagonal 
and  altitude. 

912.  It  is  required  to  construct  the  rhomboid,  having  given 
i.   The  longer  side,  sum  of  its  diagonals,  and  larger  angle 

made  by  the  diagonals. 

ii.   Its  lesser  angle,  longer  diagonal,  and  sum  of  two  ad- 
jacent sides. 

iii.  Its  lesser  angle,  longer  side,  and  sum  of  its  altitude  and 
lesser  side. 


Problems  in  Quadrilaterals  and  Circles.      151 

iv.  Its  lesser  angle,  shorter  side,  and  difference  of  its  longer 
diagonal  and  longer  side. 

913.  It  is  required  to  construct  an  isosceles  trapezoid,  having 
given 

i.  One  leg,  diagonal,  and  longer  base. 

ii.  Its  longer  base,  diagonal,  and  lesser  angle. 

ill.  Its  diagonal,  altitude,  and  leg. 

iv.  Its  longer  base,  lesser  angle,  and  sum  of  altitude  and  leg. 

914.  It  is  required  to  construct  the  trapezoid,  having  given 
i.   Its  longer  base,  lesser  angle  (i.e.  angle  formed  by  longer 

base  and  a  leg),  and  its  diagonal. 

ii.   Its  longer  base,  one  leg,  lesser  angle,  and  altitude. 

iii.   Sum  of  its  bases,  the  two  legs,  and  lesser  angle. 

iv.  Difference  of  its  bases,  the  two  legs,  and  angle  formed 
by  its  diagonals. 

III.   Circles. 

915.  Having  given  a  circle,  it  is  required  to  construct 

i.  Three  equal  circles,  tangent  to  the  given  circle  externally 
and  tangent  to  each  other. 

ii.  Three  equal  circles,  tangent  to  the  given  circle  internally 
and  tangent  to  each  other. 

iii.   Eour  equal  circles,  as  in  i  and  ii. 

iv.   Five  equal  circles,  as  in  i  and  ii. 

V.    Six  equal  circles,  as  in  i  and  ii. 

vi.  A  circle  tangent  to  three  given  circles. 

IV.   Transformation  of  Figures. 

916.  To  transform  a  given  triangle  into  an  equivalent  isos- 
celes one  having  the  same  base. 

917.  To  transform  a  given  isosceles  triangle  into  an  equiva- 
lent equilateral  one. 


152  Plane  Geometry. 

918.  To  transform  a  given  triangle  into  an  equivalent  equi- 
lateral one. 

919.  To  transform  a  given  triangle  into  another  equivalent 
triangle  whose  base  and  altitude  shall  be  equal.- 

920.  To  transform  a  given  triangle  into  another  equivalent 
triangle,  and  similar  to  a  given  triangle. 

921.  To  transform  a  given  triangle  into  a  triangle  with  one 
angle  unchanged  and  its  opposite  side  parallel  to  a  given  line. 

922.  To  transform  a  given  triangle  into  an  equivalent  tri- 
angle with  a  given  perimeter. 

923.  To  transform  a  triangle  into  a  trapezoid,  one  of  whose 
bases  shall  be  the  base  of  the  triangle,  and  one  of  its  adjacent 
angles  one  of  the  basal  angles  of  the  triangle. 

924.  To  transform  a  given  triangle  into  a  right  triangle  with 
given  perimeter. 

925.  To  transform  a  given  triangle  into  a  parallelogram 
with  given  base  and  altitude. 

926.  To  transform  a  parallelogram  into  a  parallelogram  with 
a  given  side. 

927.  To  transform  a  parallelogram  into  a  parallelogram 
having  a  given  angle. 

928.  To   transform  a  parallelogram  into   a  parallelogram 

with  given  altitude. 

929.  To  transform  a  square  into 

i.  A  right  triangle. 

ii.  An  isosceles  triangle. 

ill.  An  equilateral  triangle. 

iv.  A  rectangle  with  given  side. 

V.  A  rectangle  with  given  perimeter, 

vi.  A  rectangle  with  given  difference  of  sides. 

vii.  A  rectangle  with  given  diagonal. 


Transformation  of  Figures.  153 

930.  To  transform  a  rectangle  into 
i.   A  square. 

ii.  An  isosceles  triangle. 

iii  An  equilateral  triangle. 

iv.  A  rectangle  with  given  side. 

V.  A  rectangle  with  given  perimeter. 

vi.  A  rectangle  with  given  difference  of  sides, 

vii,  A  rectangle  with  given  diagonal. 

931.  It  is  required  to  construct  a  parallelogram  equivalent 
to  the 

i.  Sum  of  two  given  parallelograms  of  equal  altitudes. 

ii.  Difference  of  two  given  parallelograms  of  equal  altitudes. 

iii.  Sum  of  two  given  parallelograms  of  equal  bases. 

iv.  Difference  of  two  given  parallelograms  of  equal  bases. 

V.  Sum  of  two  given  parallelograms. 

vi.  Difference  of  two  given  parallelograms. 

932.  It  is  required  to  transform  a  given  parallelogram  into 
i.   A  triangle. 

ii.  An  isosceles  triangle, 

iii.  A  right  triangle. 

iv.  An  equilateral  triangle. 

v.  A  square. 

vi.  A   rhombus   having   for  a  diagonal   one   side  of  the 
parallelogram. 

vii.  A  rhombus  having  a  given  diagonal. 

viiL  A  rhombus  having  a  given  side. 

ix.  A  rhombus  having  a  given  altitude. 

X.  A  parallelogram  having  a  given  side  and  diagonal. 

933.  To  transform  a  trapezoid  into 
i.   A  triangle. 

ii.   A  square. 

iii.   A  parallelogram  having  for  one  base  the  longer  base  of 
the  trapezoid. 

iv.   An  isosceles  trapezoid. 


154  Plane  Geometry. 

934.  To  transform  a  trapezium  into 

i.   A  triangle. 

ii.   An  isosceles  triangle  with  given  base. 
ill.   A  parallelogram. 

iv.   A  trapezoid  with  one  side  and  the  two  adjacent  angles 
unchanged. 

V.     Division  of  Figures. 

935.  It  is  required  to  divide  a  given  triangle  into  any  num- 
ber of  equivalent  parts  by  lines  drawn  from  one  vertex. 

936.  It  is  required  to  divide  a  given  triangle  into  any 
number  of  equivalent  parts  by  lines  drawn  from  any  point 
in  its  perimeter. 

937.  It  is  required  to  divide  a  given  triangle  into  any  num- 
ber of  equivalent  parts  by  lines  drawn  from  any  point  in  the 
triangle. 

938.  It  is  required  to  divide  a  given  triangle  into  any  num- 
ber of  equivalent  parts  by  lines  parallel  to  one  side. 

939.  It  is  required  to  divide  a  given  triangle  into  any 
number  of  parts,  whose  areas  shall  be  in  a  given  ratio,  by 
lines  drawn  from  one  vertex. 

940.  It  is  required  to  divide  a  given  triangle  into  any  num- 
ber of  parts,  whose  areas  shall  be  in  a  given  ratio,  by  lines 
drawn  from  any  point  in  the  perimeter. 

941.  It  is  required  to  divide  a  given  triangle  into  any  num- 
ber of  parts,  whose  areas  shall  be  in  a  given  ratio,  by  lines 
drawn  from  any  point  in  the  triangle. 

942.  It  is  required  to  divide  a  given  triangle  into  any  number 
of  parts,  whose  areas  shall  be  in  a  given  ratio,  by  lines  dra^vn 
parallel  to  one  side. 


Division  of  Figures.  155 

943.  It  is  required  to  divide  a  triangle  in  the  ratio  of  m  to  n 

by  a  line  perpendicular  to  the  base. 

944.  It  is  required  to  divide  a  given  parallelogram  into  any 
number  of  equal  parts  by  lines  drawn  parallel  to  one  pair  of 
sides. 

945.  It  is  required  to  divide  a  given  parallelogram  into  any 
number  of  parts,  whose  areas  shall  be  in  a  given  ratio,  by  lines 
parallel  to  one  pair  of  sides. 

946.  It  is  required  to  divide  a  given  parallelogram  into 
two  equivalent  parts  by  a  line  drawn  from  any  point  in  the 
perimeter. 

947.  It  is  required  to  divide  a  given  parallelogram  into  two 
equivalent  parts  by  a  line  drawn  through  any  point  in  the 
parallelogram. 

948.  It  is  required  to  divide  a  given  parallelogram  into  two 
parts,  whose  areas  shall  be  in  a  given  ratio,  by  a  line  drawn 
from  one  vertex. 

949.  It  is  required  to  divide  a  given  parallelogram  into  two 
parts,  whose  areas  shall  be  in  a  given  ratio,  by  a  line  drawn 
from  any  point  in  the  perimeter. 

950.  It  is  required  to  divide  a  given  parallelogram  into  two 
parts,  whose  areas  shall  be  in  a  given  ratio,  by  a  line  drawn 
from  any  point  in  the  parallelogram. 

951.  It  is  required  to  divide  a  parallelogram  into  two  equiva- 
lent parts  by  a  line  drawn  parallel  to  a  given  line. 

952.  It  is  required  to  divide  a  parallelogram  into  two  parts, 
whose  areas  shall  be  in  a  given  ratio,  by  a  line  drawn  parallel 
to  a  given  line. 

953.  It  is  required  to  divide  a  parallelogram  into  any  number 
of  equivalent  parts  by  lines  drawn  from  either  vertex. 


156  Plane  Geometry. 

954.  It  is  required  to  divide  a  parallelogram  into  any  num- 
ber of  equivalent  parts  by  lines  drawn  from  any  point  in  its 
perimeter. 

955.  It  is  required  to  divide  a  parallelogram  into  any  num- 
ber of  equivalent  parts  by  lines  drawn  from  any  point  in  the 
parallelogram. 

956.  It  is  required  to  divide  a  parallelogram  into  any  num- 
ber of  equivalent  parts  by  lines  parallel  to  a  given  line. 

957.  It  is  required  to  divide  a  parallelogram  into  any  num- 
ber of  parts,  whose  areas  shall  be  in  a  given  ratio,  by  lines 
parallel  to  a  given  line. 

958.  It  is  required  to  divide  a  trapezoid  into  two  equivalent 
parts  by  a  line  drawn 

i.  Parallel  to  the  bases. 

ii.  Perpendicular  to  the  bases. 

ill.  Parallel  to  one  of  the  legs. 

iv.  Through  one  of  its  vertices. 

V.  Through  a  given  point  in  one  of  its  bases, 

vi.  Through  any  point  in  its  perimeter. 

vii.  Through  any  point  in  the  trapezoid, 

viii.  Parallel  to  a  given  line. 

959.  It  is  required  to  divide  a  trapezoid  into  any  number  of 
equivalent  parts  by  lines  drawn 

i.  Parallel  to  the  bases. 

ii.  Perpendicular  to  the  bases. 

iii.  Parallel  to  one  of  its  legs. 

iv.  Through  one  of  its  vertices. 

V.  Through  any  point  in  one  of  its  bases. 

vi.  Through  any  point  in  its  perimeter. 

vii.  Through  any  point  in  the  trapezoid. 

viii.  Parallel  to  a  given  line. 


Division  of  Figures.  157 

960.  It  is  required  to  divide  a  given  trapezoid  into  any 
number  of  parts  whose  areas  shall  be  in  a  given  ratio  by- 
lines drawn 

i.   Parallel  to  the  bases. 

ii.   Perpendicular  to  the  bases. 

iii.   Parallel  to  one  of  the  legs. 

iv.   Through  either  vertex. 

V.   Through  any  point  in  one  of  the  bases. 

vi.  Through  any  point  in  its  perimeter. 
vii.  Through  any  point  in  the  trapezoid. 
viii   Parallel  to  a  given  line. 

961.  It  is  required  to  divide  a  trapezium  into  two  equivalent 
parts  by  a  line  drawn  from  either  vertex. 

962.  It  is  required  to  divide  a  trapezium  into  two  equivalent 
parts  by  a  line  drawn  from  any  point  in  its  perimeter. 

963.  Given  the  diameter  of  a  circle,  it  is  required  to  con- 
struct a  straight  line  equal  to  the  circumference  in  length. 

From  654  it  is  evident  that  it  can  only  be  approximated. 


Let  AB  be  the  given  diameter  and  C  its  middle  point.  Ex- 
tend AB  indefinitely  as  AS.  Make  BD  and  DE  each  equal  to 
AB.  At  E  erect  the  perpendicular  EQ,  and  on  it  make  EF 
and  FG  each  equal  to  AB.  Join  AGy  AF,  DG,  and  DF.  Lay 
off  ER  and  HK  each  equal  to  AG,  and  from  K  lay  off  KL 
equal  to  AF.  Again,  from  L  make  LM  equal  to  i)G^,  and  MN 
equal  DF.  Bisect  EN  dX  P;  bisect  EP  at  R ;  and  then  trisect 
EB  at  T  and  W.    Then  CT  will  be  the  required  line  approxi- 


158  Plane  Geometry. 

mately  equal  to  the  circumference  of  tlie  circle  whose  diameter 
is  AB ;  for,  calling  the  diameter  unity, 

EL  =  2CH-KL  =  2Vi3  -  VlO, 
LM=VE, 

MN=^  V2. 
.-.  J5;iV=2Vi3-VlO+V5+V2, 
and     ^r=  3J^(2Vi3  -  VlO  Ji^-y/l +^), 
and  therefore 

or  =  2J  +  tV(2  Vi3  -  ViO  +  VS  +  V2)  =  3.1415922+. 

Q.E.P. 


Solutions  by  Algebra. 


159 


SOLUTIONS  BY  ALGEBRA. 


964.   To  compute  the  altitude  of  a  triangle  in  terms  of  its 


sides. 


A  c  B       D  A  D       c  B 

Let  Z^  be  an  acute  angle.     .-.  the  altitude,  designated  by 
h,  may  fall  either  without  or  within  the  triangle.     In  either 

2c 
Aff  +  h'  =  h\ 

.-.  ^2  ^  52  _  ^^ 

^g^45V-ffl4-c2-ay 
4  c2 

.    72^  (2  6c  +  6'  +  c^  -  a2)(26c  -  52  -  c^  4-  g^ . 
4c2 

,2_(b-}-c+a){h-\-c  —  a)(a-\-b  —  c)(a—b-\-c) 
*  4  c^ 

Now  let  one  half  the  sum  of  the  sides  be  designated  by  s, 
.*.  a  +  6  +  c  =  2  s, 

6  4-  c  —  a  =  2(s  —  a), 

a  +  6  —  c  =  2(s  —  c), 

a  --  &  +  c  =  2(s  -  6). 


i6o 


Plane  Geometry. 


^2  _  2  g .  2 (s  -  g)  2 (g  -  &)  .  2 (g  -  c) 

,2  _  16  s{s  —  a)(s  —  b)  (s  —  c) 

**'      ■"  4c2 

2    

•*•   «-  =  -  Vs(s  —  a)  (s  —  b)  (s  —  c). 
c 

965.   To  compute  the  area  of  a  triangle,  the  sides  only  being 
given. 


Area^J50  =  ^  =  ^.7i; 


h  =  -  Vs(s  —  a)  (g  —  6)  {s  —  c)' 
c 


.-.  Area  -450  =  ^s(s  —  a){s  —  b)  (s  —  c), 

966.   To  compute  the  medians  of  a  triangle  in  terms  of  its 

sides. 

<0 


A  c 

Designate  a  median  by  m. 

.-.  a2  +  &2  =  2m2  +  2 


a2  +  62  =  2m2-|-|. 


Solutions  by  Algebra. 


i6i 


2m  =V2(^T6^)-c^ 

967.   To  compute  the  bisector  of  an  angle  of  a  triangle  in 
terms  of  its  sides. 

Circumscribe  a  circle  about  the 
triangle  ABC.  Extend  CD,  the 
bisector  of  the  ZACB,  to  meet 
the  circumference  in  E,  and  draw 
BE. 

Jc'-\-AD'BD  =  aby 

,'.  k'  =  ab-AD-BD. 

AD:BD::b:a. 

.\  AD:b::BD:a. 

.'.  AD  -\-  BD;  AD  ::b  -h  a:b. 

,'.  AD-\-BD:a-i-b::AD:b. 

.-.  AD-\-BD:a  +  b::AD:b::BD:a; 

c         AD     BD 


or 


AD  = 


a  +  b 

be 
a  +  6 


•.  Jc'^ab- 


and  BD 


ab(? 


ac 


a-\-b 


{a  +  by 
.'.  Tc^^abfl-^    ^     V 

.   j^^ob[_{a  +  by-(?-] 
{a^by 

.   j^2  _  a5  (g  +  &  +  c)  (g  4-  ^  —  c) 
{a  +  by 


1 62  Plane  Geometry. 

,2  _  ab  '2s2(s  —  c) 
(a-^bf 

(a-hby 


.'.  k 


a-\-b 


Vabs  (s  —  c). 


968.   To  compute  the  area  of  a  triangle  in  terms  of  its  sides 
and  radius  of  the  circumscribing  circle. 

b:2r::h:a, 

.-.    ab  =  2rh. 

,-.  abc  =  2rch. 

ch 


Area  ABC = 


2' 


Area  ud-BC 


abc 
4r* 


969.   To  compute  the  radius  of  the  circumscribing  circle  of 
a  triangle  in  terms  of  its  sides. 
Sug.     Use  previous  diagram. 

ab  =  2rh. 
"         2h~Y'^"' 


h  =  -  Vs  (s  —  a){s  —  b)  (s  —  c). 
c 


/i  = 


2  Vg  (g  —  a)  (s  —  b)(s—  c) 
c 
.    cib  ^^  c 

2      2-y/s{s-a)(s-b){s-c) 


.'.  r  — 


abc 


4  Vs  {s  —  a){s^  b)  (s  —  c) 


Solutions  by  Algebra, 


163 


970.  To  compute  the 
radius  of  the  inscribed 
circle  in  terms  of  the  sides 
of  the  triangle. 


cf 


ar 


Area  ^5(7=^  +  ^  + 


hr 


Area  ABC     1 


{a-\-h-\-c)r^  _2sr^ 


=  sr. 


Vs(s  —  a)(s  —  b)  (s  —  c). 


971.  The  sides  of  a  triangle  are  480  feet,  600  feet,  and  720 
feet.     Compute 

L  its  area ;  iv.  the  three  angle  bisectors ; 

ii.  the  shortest  altitude ;        v.  radius  of  circumscribed  circle ; 
iii.  the  longest  median ;  vi.  radius  of  inscribed  circle. 

972.  The  sides  of  a  triangle  are  36, 40,  and  44  meters.  Com- 
pute the  areas  in  ares  of  the  two  triangles  formed  by  the  bisector 
of  the  angle  between  the  last  two  sides. 

973.  Two  sides  of  a  parallelogram  are  15  and  22  inches,  and 
one  diagonal  is  18  inches.  Compute  the  other  diagonal  and 
the  area. 

974.  The  sides  of  a  triangle  are  42,  35,  and  28  meters.  Com- 
pute the  area  of  the  triangle  formed  by  the  median  and  angle 
bisector  drawn  between  the  first  two  sides. 

975.  The  sides  of  a  triangle  are  54, 72,  and  80  feet.     Compute 
i.  its  area ;         ii.  the  area  of  the  circumscribed  circle ; 

iii.  the  area  of  the  inscribed  circle. 

976.  A  field  in  form  of  a  trapezium  has  one  right  angle 
(/-A) J  and  its  sides  are  AB  =  58  rods,  BC=  55  rods,  CD  =  37 
rods,  and  AD  =  32  rods.     Compute  the  area  of  the  field. 


164 


Plane  Geometry. 


SPECIAL  THEOREMS. 

977.  The  point  of  concurrence  of  the  altitudes  of  a  triangle 
is  called  the  ortJiocenter,  and  the  point  of  concurrence  of  the 
perpendicular  bisectors  of  the  sides  is  called  the  circumcenter, 
being  the  center  of  the  circumscribed  circle;  the  point  of 
concurrence  of  the  medians  is  called  the  midcenter,  and  the 
point  of  concurrence  of  the  angle  bisectors  is  called  the  incenter. 

A  complete  quadrilateral  is  the  figure  formed  by  four  straight 
lines  intersecting  one  another  in  six  points. 

978.  In  every  triangle  the  midcenter,  the  orthocenter,  and 
the  circumcenter  are  collinear;  and  the  line  joining  the  first 
two  is  twice  that  joining  the  last  two. 


Post.     Let  D  be  the  orthocenter,  E  the  midcenter,  and  F  the 
circumcenter  of  the  triangle  ABC. 

To  Prove.     DE  and  EF  are  one  line,  and  DE  =  2  EF. 

Cons.     Draw  NM. 

Dem.    NF  is  II  to  AD,  (?)    FM  is  II  to  CD,  (?)    NM  is  II  to 
AC.     (?) 

.-.  A  ACD  is  similar  to  A  FNM.     (?) 


2NM=AC.     (?) 


*  AC     2*     ^'^ 


Special  Theorems. 


165 


mi":  AD::  NM:  AC.     (?) 


AE     2      "^'^ 


AD     AG     2*     ^'^ 


NE 
AE 


NF 
AD 


NE:AE::NF:AD.  Z.  DAE  =  Z  Ji^.?^. 

.-.  A  DAE  is  similar  to  A  ENF.     (?) 

.-.  ZDEA  =  ZNEF.     (?) 
.'.  D^  and  jE7i^  are  in  the  same  line. 


(?) 


.%  NE:AE::EF:DE.     (?) 


(?) 

Q.E.D. 


979.  If  a  triangle  be  inscribed  in  a  circle,  and  lines  perpen- 
dicular to  the  sides  be  drawn  from  any  point  in  the  circum- 
ference, the  feet  of  these 
perpendiculars  are  collinear. 

Post.  Let  the  Js  be  PE, 
PF,  and  PD. 

To  Prove.  E,  F,  and  D 
are  collinear. 

Cons.  Draw  EF,  FD,  APj 
and  PB. 

Dem.  If  a  circle  be  de- 
scribed with  AP  a  diameter, 
its  circumference  will  pass 
through  E  and  F. 

Similarly,  if  a  circumfer- 
ence be  described  with  PB  a  diameter,  it  will  pass  through  F 
and  D.    (?) 

.-.  ZEAP=ZEFP',  (?) 

Z  CAP  is  sup.  to  Z  EAP ;         (?) 

Z  CAP  is  sup.  to  Z  CBP.  (?) 


1 66 


Plane  Geometry, 


ZEAP=ZCBP=ZEFP',    (?) 
Z  CBP  is  sup.  to  Z  PFD.  (?) 

Z.EFP  is  sup.  to  Z  PFD.  (?) 

EF  and  i^Z)  form  one  line. 
Ef  Fy  and  2)  are  collinear. 


Q.E.D. 


>.  If  the  middle  point  of  the  line  joining  the  orthocenters 
and  circumcenters  be  used  as  a  center  and  a  circle  described 
with  a  radius  equal  to  one  half  the  radius  of  the  circumscribed 
circle,  the  circumference  will  bisect  the  segments  of  the  alti- 
tudes between  the  orthooenter  and  the  vertices. 

O 


/ 

\ 

>i 

kI 

/  \ 

/O^ 

^  X 

/m/^'"^' 

<^5><^\ 

//     __,--"*' 

-'''' 

--o^ 

E 


^n 


Post  Let  N  be  the  orthocenter  and  0  the  circumcenter  of 
the  A  ABC.  Let  M,  P,  and  Q  be  the  mid-points  of  AN,  NO, 
and  NB  respectively. 

To  Prove.     PM=:  PG,  etc. 

Cons.    Draw  MP,  PG,  OA,  and  OB. 

Dem.  MP:=iAO  and  PG  =^  i  OB.    (?) 

AO=OB.  (?) 

.-.  MP=^PG.  (?) 


Special  Theorems. 


167 


.'.  if  a  circle  be  described  with  Pas  a  center  and  radius  PM^ 
the  circumference  will  pass  through  Gy  and  therefore  bisect 
NA  and  NB. 

In  a  similar  manner  this  circumference  will  bisect  NO- 

Q.E.D. 

981.  The  circumference  which  bisects  the  segments  of  the 
altitudes  of  a  triangle  between  the  orthocenter  and  the  ver- 
tices, bisects  also  the  sides  and  passes  through  the  feet  of  the 
altitudes.     (This  is  called  the  nine-point  circle.) 


Post.    Let  OR  be  the  line  joining  the  orthocenter  and  cir- 
cumcenter  and  Q  its  center. 

To  Prove.     That  the  circumference  described  with  Q  as  a 

AR 

center  and  -— -  as  a  radius  will  pass  through  points  D,  E,  Fy 

H,  K,  and  N. 

Cons,    Draw  FS  II  CO,  and  EP II  BO. 

Join  JIP  and  ^/S. 

Dem, 


FS  is  II  to  BE. 

(?) 

ES  is  II  to  PO. 

(?) 

RFis  II  toPO. 

(?) 

•.  ES  is  II  to  RF. 

(?) 

i68 


Plane  Geometry, 


BF=PO.  (?) 

.-.  APOQ  =  AFBQ.  (?) 

.-.  PF  bisects  OE,  i.e.  passes  through  point  Q. 

.'.  PQ  =  QF.     (Homol.  sides  of  equal  A.) 

The  ±  bisectors  of  HF,  KN,  and  DE  pass  through  Q  (422). 

.-.  Q  is  equidistant  from  points  D,  E^  F,  H,  K,  iV,  and  the 
mid-points  of  OA,  OB,  and  00. 

.-.  the  circumference  which  passes  through  one  will  pass 
through  all.  q.e.d. 

982.  The  circumference  of  the  nine-point  circle  bisects 
every  line  drawn  from  the  orthocenter  to  the  circumference 
of  the  circumscribed  circle. 


Post.  Let  0  be  the  center  of  the  nine-point  circle,  and  Fff 
any  chord  of  the  circumscribed  circle  passing  through  the 
orthocenter  D. 

To  Prove.     DN=  NH  and  DQ  =  FQ. 

Cons.  Draw  OQ,  OJSf,  LH,  and  LF^  and  draw  OR  to  the 
mid-point  of  DH. 


Special  Theorems. 


169 


Dem,        OQ  =  iFL    (?)     and     ON=\LH.  (?) 

.♦.  OQ=ON.  (?) 

OM  is  II  to  LH,  and  OE=i  LH.  (?) 

.-.  OR  =  OK  (?) 

.-.  on  coincides  with  ON.  (?) 

.-.  JVis  the  center  of  DH.  (?)       q.e.d. 

983.  If  a  transversal  cuts  the  sides  of  a  triangle,  produced 
if  necessary,  the  product  of  three  non-adjacent  segments  of 
the  sides  is  equal  to  the  product  of  the  other  three  segments. 

A 

F  Bj 

Post.  Let  ABC  be  a  triangle,  and  DFH  a  transversal  inter- 
secting the  sides. 

To  Prove.      BH -  FC -  AD=  HC -  AF -  DB. 
Cons.    Draw  CK  II  to  DB. 

Dem.  KC  is  II  to  DB.  (?) 

.-.  BH'.  CH'.'.DB'.CK,  (?) 

and  FC'.AF'.-.CK'.AD.  (?) 

.-.  BH'FC.CH'AF'.iDB'.AD.  (?) 

.♦.  BH'FC'AD=CH'AF'  DB.  (?)       q.e.d. 

(This  is  Menelaus's  Theorem,  discovered  about  80  b.c,  and 
the  following  is  its  converse.) 

984.  If  three  points  be  given  in  the  sides  of  a  triangle,  or 
the  sides  extended,  so  that  the  product  of  three  non-adjacent 
segments  is  equal  to  the  product  of  the  other  three,  the  three 
points  are  collinear. 


170 


Plane  Geometry. 


Post.  Let  ABC  be  a  triangle,  and  Z>,  F,  and  H  be  three 
points  so  chosen  that  BH-  FC-  AD^CH-AF-  DB. 

To  Prove.     D,  F,  and  H  are  collinear. 

Cons.  Draw  IfF  and  extend  it  to  meet  AB  in  some  point 
as  K. 

Dem.  BH'FC'AK=  CH  -  AF  -  KB. 

BH'FC'  AD  =  CH'  AF  -  DB. 
AK^KB 

' '  AD     DB 
.-.  AK'DB  =  AD'KB. 

.'.  points  D  and  ^must  coincide,  and 

.-.  D,  Fj  and  H  are  collinear.  q.e.d. 

985.  The  mid-points  of  the  three  diagonals  of  a  complete 
quadrilateral  are  collinear. 


(?) 
(?) 

(?) 


Post.  Let  ABFDE  be  a  complete  quadrilateral,  L,  M,  and 
M  being  the  mid-points  of  the  three  diagonals  AC,  BD,  and  EF, 
joined  by  the  lines  LM  and  MN. 


Special  Theorems.  171 

To  Prove.    L,  M,  and  N  are  collinear. 

Cons.     Draw  NO  II  to  ED,  join  G^Jf  and  LH. 

Dem.     GM  is  W  to  BF.     (?) 

.-.  GJf  bisects  CD.     (?) 

Similarly,  LH  is  II  to  AF  and  bisects  CD. 

.-.  LH,  MO,  and  ED  are  concurrent  in  point  K. 

E-egarding  AE  as  a  transversal  of  the  triangle  CDF, 

AD  '  BF-  EC=  AF  '  BC  '  ED.      (?) 

AD    BF    EC^AF    BCED^       /^s 

'**     2    *    2    *    2         2    '    2    '    2   *      ^'^ 

.-.  LK'  MO'  NH=LH  -  MK  •  iV^(7.     (?) 

.-.  L,  M,  and  N  are  collinear.     (?)  q.e.d. 


If  lines  drawn  from  the  vertices  of  a  triangle  are  con- 
current, the  product  of  three  non-adjacent  segments  of  the  sides 
equals  the  product  of  the  other  three. 


Post.    Let  ABC  be  a  triangle,  0  any  point  in  its  plane  through 
which  pass  the  lines  AQ,  CN,  and  BH. 

To  Prove.       BQ  -  HC  -  AN=  QC  -  AH  -  NB. 

Dem.  BQ.-HC-AO  =  BC'AH'  QO. 

(BH  is  a  transversal  cutting  A  ACQ.) 

qC '  AG  '  NB==BC '  QO  '  NA. 

(JSfC  is  a  transversal  cutting  A  ABQ.) 


172  Plane  Geometry. 

"  qC'NB     NA      ^'^ 


.-.  BQ'HG'NA^qC^NB'  AH,     (?)  q.e.d. 

(This  is  Ceva's  theorem,  discovered  in  1678,  and  the  follow- 
ing is  its  converse.) 

987.  If  three  points  are  given  on  the  sides  of  a  triangle,  or 
the  sides  extended  so  that  the  product  of  three  non-adjacent 
segments  of  the  sides  equals  the  product  of  the  other  three, 
the  lines  joining  these  points  with  the  opposite  vertices  are 
concurrent. 

A 


Post.    Let  ABO  be  a  triangle  and  N,  Q,  and  ff  three  points 
in  the  sides  such  that  BQ-HC-AN^QG-  AH  •  NB. 
To  Prove.     AQ,  BHj  and  OiVare  concurrent. 
Gons.     Extend  BO  to  AG,  meeting  it  in  point  W. 

Dem. 


BQ 

.WO'A]Sr=:QG-AW'BN. 

(?) 

BQ 

.  HO .  A]Sr=  qO'AH'NB. 

(?) 

WO     AW 
"HO     AH 

(?) 

.^ 

WO'AH=AW'HO. 

(?) 

.•.  points  H  and  W  coincide. 

.*.  AQ,  BH,  and  O^are  concurrent.    *  q.e.d. 


Special  Theorems.  173 

If  a  circumference  intersect  the  sides  of  a  triangle,  the 
product  of  the  three  non-adjacent  secants  and  their  external 
segments  equals  the  product  of  the  other  three  and  their  external 
segments. 


Post.  Let  AKD  be  a  triangle  with  its  sides  intersected  by 
a  circumference  OHP. 

To  Prove.     (AB  -  AC)  (DF  -  DH)  {KN  •  KG) 

=  {DC '  DB)  (KH .  KF)  (AO  •  AIT). 

Dem.  AB'AC=AO'AN.      (?) 

DF'DH=DG'DB.      (?) 

KN'KO^KH'ET.      (?) 

.'.  {AB  .  AC)  {DF .  DH)  {KN  -  KO) 

=  {DC '  DB)  {KH .  KF)  {AO  •  AJP).     (?)       q.e.d. 

(This  is  kaown  as  Carnot's  Theorem.) 

989.  If  the  opposite  sides  of  an  inscribed  hexagon  intersect 
each  other  the  points  of  intersection  are  collinear. 

Post.  Let  A.BD  be  an  inscribed  hexagon  with  sides  BA 
and  ED  meeting  at  P,  CD  and  AF  at  Q,  and  BC  and  FE 
at  R. 

To  Prove.     P,  Q,  and  B  are  collinear. 

Com.  Extend  BF  to  PB  at  L,  and  extend  QC,  and  PB 
until  they  meet  at  M. 


174 


Plane  Geometry. 


M 


Dem.     Eegarding  the  triangle  NLM. 

LP '  MD '  NE  =  PM '  DN  •  EL,        (Menelaus) 
MQ'NF'LA^QN'FL'  AM, 
NR'LB'MC  =  RL'BM'  CN. 
.-.  (LP .  MQ '  J^E)  (MD  .  MG)  (NE  -  NF)  (LA  -  LB)  = 

(PM-  QN'  ML)  (DN-  CN)  (EL  -  FL)  (AM-  BM) 
(DM '  CM)  (EN .  FN)  (AL  •  BL)  = 
(DN .  CN)  (EL  .  FL)  (AM  •  BM).  (Carnot's  Theorem.) 
.-.  LP'MQ'NE==PM'  QN'RL.     (?) 
.'.  P,  Q,  and  E  are  collinear.  (?)         q.e.d. 

(This  theorem  was  discovered  by  Pascal  in  1639,  when  he 
was  16  years  old,  and  is  therefore  known  as  Pascal's  Theorem.) 

990.   The  bisectors  of  adjacent  exterior  and  interior  angles 
of  a  triangle  divide  the  opposite  side  harmonically. 


Dem. 


F  B        D 

BD:DA::CB:CA     (?) 
FB:FA::CB:CA     (?) 


Special  Theorems. 

BD:DA::FB:FA     (?) 


^75 


.'.  BD  is  divided  harmonically.  q.e.d. 

Co7\     CA  and  CB  divide  FD  harmonically,  for,  taking  the 
last  proportion  by  alternation, 

BD:FB::DA:  FA. 

Consequently  the  four  points,  Fj  B,  D,  and  Aj  are  called  har- 
monic points,  and  the  two  pairs,  A  and  B,  and  F  and  D,  are 
called  conjugate  harmonic  points.  Likewise  the  four  lines,  CF, 
CB,  CD,  and  CA,  are  together  called  an  harmonic  pencil,  and 
each  line  is  called  an  harmonic  ray. 

991.   In  a  complete  quadrilateral,  each  diagonal  is  divided 
harmonically  by  the  other  two. 


Dem.     In  A  FAE  the  trans.  DM  gives 

DF'AB'  EM=  AD 'BE'  MF, 
DF'AB'EL  =AD'BE'  LF, 


(?) 
(?) 

(?) 


.•.  MF  is  divided  harmonically. 

In   like  manner   it  may  be  shown  that  AL  and  MD  are 
divided  harmonically.  q.e.d. 


.-. 

EM     MF 
EL      LF 

EM 

EL  .-.MF:  LF. 

EM 

MF'.'.EL  :  LF. 

176  Plane  Geometry. 


MAXIMA  AND   MINIMA. 

992.  Among  quantities  of  the  same  kind,  that  which  is 
greatest  is  called  the  maximum  (plural,  maxima),  and  that 
which  is  the  smallest  is  called  the  minimum  (plural,  minima). 
For  example:  of  all  lines  inscribed  in  the  same  circle,  the 
diameter  is  the  greatest,  and,  therefore,  is  a  maximum;  and  of 
all  lines  drawn  from  the  same  point  to  a  given  straight  line, 
that  which  is  perpendicular  is  the  shortest,  and  is,  therefore, 
a  miriimum. 

If  two  or  more  plane  figures  have  equal  perimeters,  they  are 
said  to  be  isoperimetric. 

A  plane  figure  is  said  to  be  a  maximum  or  a  minimum  when  its 
area  is  a  maximum  or  a  minimum. 

THEOREMS. 

993.  If  any  ntiinber  of  triangles  have  the  same  or  equal  bases  and 
equal  areas,  that  which  is  isosceles  has  the  minimum  perimeter. 

^^  Post.  Let  ABC  and  ABD  be  two 
triangles  having  the  same  base  AB 
and  equal  areas,  and  let  ACB  be 
isosceles,  having  CA  equal  to  CB. 

To  Prove. 
AC-{-CB-{-AB<AD+DB-{-AB, 
/>^         N.  \  i       or  since        AB  =  AB, 

^A^                 XM  To  Prove. 

j^^:^ ^B  AC-i-CB<AD+DB. 

Cons.  From  B  construct  a  perpendicular  to  AB,  and  extend 
it  to  meet  AC  extended  in  K.  Join  DK  and  draw  CH  through 
point  D. 

Dem.  How  must  the  altitudes  of  the  two  A  ACB  and  ADB 
compare  ? 

What  must  be  the  position,  then,  of  the  line  CH  relative 
to  AB? 


Maxima  and  Minima.  177 

How,  then,  do  the  two  A  HCB  and  CBA  compare  ?    Why  ? 

The  two  A  KCH  and  CAB  ?     Why  ? 

How,  then,  must  the  two  A  KCH  and  HCB  compare  ?    Why  ? 

What  is  the  position  of  CH  relative  to  KB  ?     Why  ? 

How,  then,  must  CK  and  CB  compare  ?     DK  and  DB  ? 

Now  compare  AC  -\-  CK  with  AD  +  DKy  and  the  pupil 
should  be  able  to  write,  or  give  orally,  a  complete  and  accurate 
demonstration  of  this  theorem. 

994.  If  any  number  of  triangles  have  equal  areas,  that  which  is 
equilateral  has  the  minimum  perimeter. 

995.  If  any  number  of  triangles  have  the  same  or  equal  bases  and 
equal  perimeters,  that  which  is  isosceles  is  the  maximum. 

Sug.  Through  the  vertex  of  the  isosceles  triangle  draw  a 
line  parallel  to  the  base.  Then  prove  that  the  vertex  of  the 
other  triangle  cannot  fall  on  that  line.  Then  compare  their 
altitudes,  and  consequently  their  areas. 


I.   If  any  number  of  triangles  be  isoperimetrio,  that  which  is 
equilateral  is  the  maximum. 

997.  If  any  number  of  triangles  have  two  sides  in  each  respectively 
equal,  that  in  which  these  sides  are  perpendicular  to  each  other  is  the 
maximum. 

Sug.  Place  them  so  that 
one  set  of  equal  sides  shall 
coincide,  as  AB.  Then 
compare  their  altitudes  PK, 
CAj  and  HN,  and  conse- 
quently their  areas. 

998.  If  any  number  of  equivalent  parallelograms  have  the  same  or 
equal  bases,  the  perimeter  of  that  which  is  rectangular  is  the  minimum. 

999.  Of  all  rectangles  of  given  area,  the  perimeter  of  a  square  is  a 
minimum. 


178 


Plane  Geometry. 


1000.  If  any  number  of  triangles  have  tlie  same  or  equal  bases  and 
tbe  same  or  equal  altitudes,  the  perimeter  of  that  which  is  isosceles  is  a 
minimum! 

1001.  Every  closed  plane  figure  of  given  perimeter  whose  area  is 
a  maximum,  must  be  convex. 

Post.  Let  ACBN  be  a  plain  concave  figure,  with,  the  straight 
line  AB  which  joins  two  of  its  points  in 
its  perimeter  lying  without. 

To  Prove.     That  ACBN  cannot  be  a 
maximum. 

Dem.  Conceive  the  figure  CAB  to  be 
revolved  about  AB  as  an  axis  till  it 
comes  to  the  position  AC'B.  Then  the 
figures  ACBN  said  AC'BNhsive  equal  perimeters,  but  the  area 
of  the  latter  will  exceed  that  of  the  former.  Hence  ACBN 
cannot  be  a  maximum  among  isoperimetrical  figures.  But 
ACBN  is  any  concave,  i.e.  non-convex,  plane  figure.  There- 
fore, of  all  isoperimetrical  plane  figures,  the  maximum  must  be 
convex. 


1002.   Of  all  plane  figures  that  are  isoperimetric,  that  which  is  a 
circle  is  the  maximum. 

Dem.  I.  It  is  evident  that  with  a  given  perimeter,  an  in- 
definite number  of  figures  of  different  shapes  and  areas  may  be 
constructed.  It  is  also  evident 
that  we  can  diminish  the  area 
indefinitely,  but  cannot  thus  in- 
crease it.  Consequently,  there 
must  be  among  all  these  figures 
having  the  same  perimeter 
either  one  maximum  figure,  or 
several  maximum  figures  of  dif- 
ferent forms  and  equal  areas. 

II.   Let  ACBFG  be  a  maxi- 
mum figure  with  a  given  per- 


Maxima  and  Minima.  179 

imeter;  then  by  1001  it  must  be  convex.  Let  also  the  line 
AB  divide  the  perimeter  into  halves ;  then  it  must  also  divide 
the  area  into  halves.  For  suppose  one  of  the  parts,  as  AFBj 
to  be  greater  than  the  other,  and  conceive  this  part  to  be 
revolved  on  AB  as  an  axis  until  it  comes  into  the  same  plane 
with  ACB,  and  let  AF'B  be  its  position  after  revolution. 
Hence  the  perimeter  of  the  figure  AF'BEGA  is  equal  to  that 
of  the  figure  AGBFGAy  but  the  area  of  the  former  is  greater 
than  that  of  the  latter.  Therefore  the  figure  ACBFG  cannot 
be  a  maximum.  But  by  hypothesis  it  is  a  maximum.  Hence 
AB  must  bisect  the  area  of  ACBFG. 

Since  ACBFG  is  a  maximum,  and  AB  bisects  the  area,  it 
follows  that  the  figure  AF'BFG  is  also  a  maximum. 

Again,  let  F  be  any  point  in  BEG  A  selected  at  random,  and 
F'  its  position  after  revolution.  Join  FF'j  FB,  FA,  F'B,  and 
F'A.  Then  AF=AF',  and  FQ  =  F'Q.  Hence,  the  two  tri- 
angles AQF  and  AQF'  are  equal. 

Therefore  FF'  is  perpendicular  to  AB. 

Similarly,  the  two  triangles  AF'B  and  AFB  are  equal. 

The  triangle  AFB  must  be  a  maximum,  otherwise  its  area 
could  be  increased  without  increasing  its  perimeter;  i.e.  with- 
out increasing  the  lengths  of  the  two  chords  AF  and  FB,  which 
would  consequently  leave  the  areas  of  the  two  segments  AGF 
and  FEB  unchanged,  and  therefore  make  up  an  area  greater 
than  ABEG,  by  which  it  is  evident  that  ACBFG  could  not  be 
a  maximum;  but  this  also  conflicts  with  the  hypothesis  which 
grants  that  ACBFG  is  a  maximum.  Consequently  the  triangle 
AFB  must  be  a  maximum,  and  the  angle  AFB  must  be  a  right 
angle.  (See  Theorem  997.)  But  F  is  any  point  in  the  curve 
BEFGA.  Hence  BFA  must  be  a  semicircle,  as  also  ACB. 
Hence  the  whole  figure  ACBFG  must  be  a  circle.  q.e.d. 

1003.  Of  all  plane  figures  having  equal  areas,  the  perimeter  of  that 
which  is  a  circle  is  the  minimum. 

Post.  Let  (7  be  a  circle,  and  A  any  other  plane  figure  having 
the  same  area. 


i8o 


Plane  Geometry. 


To  Prove.  Perimeter  G  <  perimeter  A. 

Dem.  Eor  let  J5  be  a  circle  having  a  perimeter  equal  to  that 
of  A.  Hence  by  1002  area  B  is  greater  than  area  A,  and  hence 
greater  than  area  C.  Hence  if  area  G  is  less  than  area  J5, 
what  must  be  true  of  their  perimeters,  ^.e.  their  circumferences  ? 
But  by  construction 

Perimeter  B  =  perimeter  A, 

Hence  perimeter  of  (7  is  a  minimum.  q.e.d. 

1004.  Of  all  mutnally  equilateral  polygons,  that  which  can  be  in- 
scribed in  a  circle  is  the  maximum. 


Post  Let  ABGDH=  P,  and  A'B'G'D'H'  =  P',  be  two  mutu- 
ally equilateral  polygons,  having  AB  equal  to  A'B',  BG  equal 
to  B'G'f  etc.,  of  which  ABGDH  can  be  inscribed  in  a  circle, 
and  let  N  be  the  center  of  such  circle. 

Gons.  With  the  radius  NB  construct  the  arcs  A'B\  B'G', 
CD',  etc. 


Maxima  and  Minima.  i8i 

Dem.   The  are  AB  —  arc  AB\  and  arc  50=  arc  5'C",  etc. 
Hence  circumference  ABGDR—  sum  of  the  arcs  A!B\  B'G\ 
etc. 

Hence  perimeter  of  >S  =  perimeter  of  S\ 

Therefore  Area  S  >  area  aS".  (Theorem  1002.) 

But  the  corresponding  segments  are  equal.     Hence,  subtract- 
ing their  respective  sums  from  the  above  inequality  leaves 

Area  P  >  area  P.  q.e.d. 

1005.   Of  all  isoperimetric  polygons  of  the  same  number  of  sides, 
that  which  is  equilateral  is  a  maximum. 

Post.     Let  ABCDHK  be  the  maximum  of  all  isoperimetrical 
polygons  of  any  given  number 
of  sides. 

To  Prove.    That  it  is  equilat-    w-^,^- 
eral ;  i.e.  that  '      ^ 

KH=HD  =  DCy  etc. 

Cons.  Connect  any  two  al- 
ternate vertices,  as  Aff. 

Dem.  The  A  AKH  must  be 
a  maximum  of  all  isoperimetri- 
cal triangles  having  the  common 

base  AH'j  otherwise  another  triangle,  as  ANH,  could  be  con- 
structed, having  the  same  perimeter  and  a  greater  area,  in 
which  case  the  area  of  the  polygon  ABCDHNvf ovXdi  be  greater 
than  that  of  ABCDHK.  Hence  the  latter  would  not  be  a 
maximum.  This  result  conflicts  with  our  hypothesis,  which 
grants  that  it  is  a  maximum.  Therefore  the  triangle  AKH 
must  be  a  maximum,  and  consequently  isosceles.  (See  Theo- 
rem 995.) 

Hence  AK==  KH 

Similarly,  by  joining  KD,     KH=  HD,  etc 

Hence  the  polygon  is  equilateral.  q.e.d. 


1 82  Plane  Geometry. 

1006.  The  maximum  of  all  equilateral  polygons  of  the  same  number 
of  sides  is  that  which  is  regular. 

1007.  Of  all  polygons  having  the  same  number  of  sides  and  equal 
areas,  the  perimeter  of  that  one  which  is  regular  is  a  minimum. 

Post.  Let  P  be  a  regular  polygon,  and  M  any  irregular  poly- 
gon having  the  same  number  of  sides 
and  same  area  as  P;  and  let  JV  be  a 
regular  polygon  having  the  same  number 
of  sides  and  isoperimetrical  wi^h  M. 
Dem.        Area  M  <  N. 

(See  1005  and  1006.) 
Area  M=  area  P.     Why  ? 
.-.  Area  P  <  N. 

But  of  two  regular  polygons  of  the  same  number  of  sides, 
that  which  has  the  less  area  must  have  the  less  perimeter  ? 
Why? 


jsr 


Hence  Perimeter  P  <  perimeter  N. 

and  .-.  Perimeter  P  <  perimeter  M. 

Hence  the  perimeter  of  P  is  a  minimum.  q.e.d. 

1008.  Of  all  isoperimetrio  regular  polygons,  that  which  has  the 
greatest  number  of  sides  is  the  maximum. 

Post.  Let  ABC  be  a  regular  trigon,  and  P  a  regular  tetror 
gon,  having  equal  perimeters. 

To  Prove.    Area  P  >  area  ABC. 


Maxima  and  Minima. 


183 


^  X>  B 

Cons.     Draw  from    C  any   line  CD  to  AB.     At   C  make 

Z  i)(7iT  equal  to  the  Z  CDB,  and  CH  equal  to  i>5,  and 
join  HD. 

Dem.  ACDH=ACDB.              Why? 

Hence  Avea.  ABC  =  Sivesi  AD  HC.    Why? 

But  Area  F  >  area  ADHC.    Why  ? 

Hence  Area  P  >  area  ABC. 

Similarly,  P  could  be  shown  to  be  less  than  an  isoperimetric 
regular  pentagon,  etc.  q.e.d. 

1009.  The  area  of  a  circle  is  greater  than  the  area  of  any  polygon 
of  equal  perimeter. 

1010.  Of  all  regular  polygons  having  a  given  area,  the  perimeter  of 
that  which  has  the  greatest  number  of  sides  is  a  minimum. 


Post.     Let  Q  and  P  be  two  regular  polygons  having  equal 
areas,  and  Q  having  the  greater  number  of  sides. 


184  Plane  Geometry. 

To  Prove.     That  the  perimeter  of  P  is  greater  than  that  of  Q. 
Dem.     Let  J?  be  a  regular  polygon  whose  perimeter  is  equal 
to  that  of  Qy  but  the  number  of  sides  the  same  as  P. 

Then  Q>R.         Why  ? 

But  Area  Q  =  area  P. 

.'.  Area  P  >  area  i2. 

.-.  Perimeter  P  >  perimeter  P.    Why  ? 

But  Perimeter  P  =  perimeter  Q. 

Perimeter  P  >  perimeter  Q. 

Hence  perimeter  Q  is  a  minimum.  q.e.d. 

1011.  The  circumference  of  a  circle  is  less  than  the  perimeter  of 
any  polygon  of  equal  area. 

1012.  The  rectangle  formed  by  the  two  segments  of  a  linQ  is  a 
maximum  when  the  segments  are  equal. 


Symmetry. 


i8s 


SYMMETRY. 


1013.  Two  points  are  said  to  be  symmetrical  with  respect  to 
a  point  when  they  are  equidistant  from,  and  in  the  same  line 
with,  this  point.     This  point  is  called  the  center  of  symmetry. 

1014.  Two  points  are  said  to  be  symmetrical  with  respect  to 
a  line  when  the  line  that  joins  them  is  perpendicular  to,  and 
bisected  by,  this  line.     This  line  is  called  an  axis  of  symmetry. 

1015.  Two  points  are  said  to  be  symmetrical  with  respect  to 
a  plane  when  the  line  that  joins  them  is  perpendicular  to,  and 
bisected  by,  this  plane.  This  plane  is  called  a  plane  of  sym- 
metry. 

1016.  The  distance  of  either  of  two  symmetrical  points  from 
the  center  of  symmetry  is  called  the  radius  of  symmetry. 

1017.  Two  plane  figures  are  symmetrical  with  respect  to  a 
center  J  axis^  or  a  plane  when  any  point  in  either  figure  selected 
at  random  has  a  corresponding  symmetrical  point  in  the  other. 


\ 


\C 


\ 


2f 


Fig.  I. 


S 
Fig.  II. 


M 


Fig.  III. 


Fig.  IV. 


i86 


Plane  Geometry. 


M 


N 


Fig.  V. 


Fig.  VI. 


Thus,  in  Pigs.  I.,  II.,  and  III.  the  lines  AB  and  QD  are  sym- 
metrical with  respect  to  the  center  C,  the  axis  KS,  and  the 
plane  JOT,  respectively.  In  Tigs.  IV.,  V.,  and  VI.  the  same  is 
true  of  the  triangles  ABQ  and  HB  W. 

1018.    A  plane  figure  is  symmetrical 

i.  when  it  can  be  divided  by  an  axis  into  two  figures  sym- 
metrical with  respect  to  that  axis. 

ii.  when  it  has  a  center  such  that,  if  a  line  be  drawn  through 
it  in  any  direction  at  random,  the  two  points  at  which  it 
intersects  the  perimeter  are  symmetrical  with  respect  to 
that  center. 


Thus  figure  ABCDH  is  symmetrical  with  respect  to  the 
axis  KS,  and  ABQDHK  with  respect  to  the  center  (7.  In 
the  latter  case  PP^  or  NN^  is  called  a  diameter  of  symmetry. 
(See  1016.) 


Symmetry.  187 

THEOREMS. 

1019.  The  center  of  a  circle  is  a  center  of  symmetry. 

1020.  The  diameter  of  a  circle  is  an  axis  of  symmetry. 

1021.  The  line  which  bisects  the  vertical  angle  of  an  isosceles  tri- 
angle is  an  axis  of  symmetry. 

1 022.  Either  altitude  of  an  equilateral  triangle  is  an  axis  of  symmetry. 

1023.  A  segment  of  a  circle  is  a  symmetrical  figure. 

1024.  That  part  of  a  circle  included  between  two  parallel  chords  is 
a  symmetrical  figure. 

1025.  The  common  part  of  two  intersecting  circles  is  a  symmetrical 
figure. 

1026.  The  diagonal  of  a  square  is  an  axis  of  symmetry. 

1027.  Every  equilateral  tetragon  is  a  symmetrical  figure. 
(How  many  axes  of  symmetry  does  a  square  have  ?) 

1028.  The  point  of  intersection  of  the  diagonals  of  a  parallelogram 
is  a  center  of  symmetry. 

1029.  An  isosceles  trapezoid  is  a  symmetrical  figure. 

1030.  If  one  diagonal  of  a  tetragon  divides  it  into  two  isosceles 
triangles,  the  other  diagonal  is  an  axis  of  symmetry. 

1031.  The  bisector  of  an  angle  of  a  regular  polygon  is  an  axis  of 
symmetry. 

1032.  The  perpendicular  bisector  of  one  side  of  a  regular  polygon  is 
an  axis  of  symmetry. 

1033.  If  the  angles  at  the  extremities  of  one  side  of  an  equilateral 
pentagon  be  equal,  the  pentagon  is  a  symmetrical  figure. 


i88 


Plane  Geometry. 


1034.  If  two  diametrically  opposite  angles  of  an  equilateral  hexagon 
are  equal,  the  hexagon  is  a  symmetrical  figure. 

1035.  Every  equiangular  tetragon  is  a  symmetrical  figure. 

1036.  If  a  figure  have  two  axes  of  synmietry  perpendicular  to  each 
other,  their  intersection  is  a  center  of  symmetry. 

Post.  Let  ABDH,  etc.,  be  a  figure  having  two  axes  of 
symmetry  PP'  and  iOf'  _L  to  each  other. 

To  Prove.  That  their  point  of  intersection  0  is  a  center  of 
symmetry. 

Cons.  From  any  point  in  the  perimeter  selected  at  random, 
as  Q,  construct  QO  J.  to  MM',  QF  l.to  PP',  and  join  TL,  OC, 
and  FG. 

Dem.  OT=:TQ  =  La  Why  ? 


Then  what  kind  of  a  figure  is  OTLC?  What  relation,  then, 
between  TL  and  OG  ?  Compare  in  a  similar  manner  TL  and 
GF.     Finally  compare  OG  and  GF. 

Hence  points  0  and  F  are  in  the  same  line  with  and  equi- 
distant from  G.     Hence  C  is  a  center  of  symmetry.  q.e.d. 


Tables. 


189 


1037-    METRIC  AND  COMMON  MEASURES  OF  LENGTH 
AND   SURFACE. 

Linear  Measuee. 
12    inches  (in.)  =  1  foot  (ft.). 

3    feet  =  1  yard  (yd.). 

5^  yards,  or  16|-  feet  =  1  rod  (rd.). 

320    rods,  1760  yai-ds,  or  5280  feet  =  1  mile  (mi.). 

Measures  of  Surface. 

144    square  inches  (sq.  in.)  t=  1  square  foot  (sq.  ft.). 
9    square  feet  =  1  square  yard  (sq.  yd. 

\0X  snnn.rp.  va.rrls.  nv  1  .  . 


30J  square  yards,  or 
272J  square  feet 
160    square  rods,  or 

10    square  chains 
640    acres 


square  yard  (sq.  yd.). 
[         =  1  square  rod  (sq.  rd.). 

=  1  acre  (A.). 

=  1  square  mile  (sq.  mi.). 


Divisions. 


The  unit. 


Multiples. 


Linear  Measure. 
A  millimeter  (mm.)  =  .001  of  a  meter. 


A  centimeter  (cm.) 
A  decimeter 
A  meter  (m.). 
A  dekameter 
A  hektometer 
A  kilometer  (km.) 


=  .01  of  a  meter. 
=  .1  of  a  meter. 

=  10  meters. 
=  100  meters. 
=  1000  meters. 


Measures 

A  square  millimeter  (qmm.) 

A  square  centimeter  (qcm.) 

A  square  decimeter 

A  square  meter  (qm.) 

A  square  dekameter 

A  square  hektometer 

A  square  kilometer  (qkm.) 


OF  Surface. 

=  .000001  of  a  square  meter. 

=  .0001  of  a  square  meter. 

=  .01  of  a  square  meter. 

=  principal  unit. 

=  100  square  meters. 

=  10000  square  meters. 

=  1000000  square  meters. 


I  go  Plane  Geometry. 

In  the  measurement  of  land,  the  square  dekameter  is  called 
an  are  (a.),  and  the  square  hektometer  is  called  a  hektare  (ha.). 

Tables  of  Equivalents. 
Length. 


Meter  = 


39.37043  in.  Inch  =  2.53998  cm. 

[  1.09362  yd.  Yard  =  0.91439  m. 

Kilometer  =  0.62138  mi.  Mile  =  1.60933  km. 

Surface. 

Q  ivr  f     _  1 1550.031  sq.  in.      Square  inch  =  6.45148  qcm. 

bquare  Meter  _  |  ^^gg^^  ^^^  ^^^      g^^^^.^  ^^^^  ^  0.83611  qm. 

Hektare  =  2.47110  A.  Acre  =  0.40468  ha. 

Approximate  Equivalents. 

Meter  =  1.1  yd.  Yard  =  .9  m. 

Kilometer  =  |  mi.  Mile  =  1.6  km. 

Square  meter  =  li  sq.  yd.  Square  Yard  =  f  qm. 
Hektare  =  2^  A.  Acre  =  |  ha. 


ENTRANCE  EXAMINATION  PAPERS. 


HARVARD  — 1899. 


New  Method.  —  The  University  provides  a  Syllabus.  The  order  of  propositions  belong- 
ing to  one  and  the  same  Book  is  not  prescribed  ;  but  it  is  not  expected  that  a  proposition  of 
a  given  Book  shall  be  proved  by  the  aid  of  propositions  appearing  in  later  Books.  Should 
the  candidate,  however,  have  used  a  text-book  in  which  the  division  into  Books  is  incon- 
sistent with  the  division  of  the  Syllabus,  and  should  he  prefer  to  follow  the  order  of  propo- 
sitions with  which  he  is  familiar,  he  will  be  allowed  to  do  so  on  stating  in  his  examination 
book  the  name  of  the  text-book  he  has  used.    Omit  one  of  the  starred  questions. 

*  1.  Prove  that  if  two  right  triangles  have  the  hypothenuse  and  a 
side  of  one  respectively  equal  to  the  hypothenuse  and  a  side  of  the 
other,  the  triangles  are  equal. 

Two  triangles  have  an  angle,  the  opposite  side,  and  another  side  of 
one  equal  respectively  to  an  angle,  the  opposite  side,  and  another  side 
of  the  other.  Are  these  triangles  necessarily  equal  ?  Give  roughly, 
by  the  aid  of  a  figure,  the  reason  for  your  answer  without  giving  a 
formal  proof. 

*  2.  Prove  that  two  parallel  straight  lines  intercept  equal  arcs  on  a 
circumference. 

A  straight  line  joins  the  centers  of  two  circles  and  cuts  the  first 
circle  in  the  points  A  and  B,  and  the  second  circle  in  the  points  C 
and  D.  Four  parallel  lines  pass  through  A,B,  C,  and  D  respectively. 
One  half  of  the  first  circumference  lies  between  the  first  two  of  these 
lines.  What  part  of  the  second  circumference  lies  between  the  second 
two  lines,  and  what  angle  do  these  four  lines  make  with  the  line  AB'i 

*  3.  A  set  of  circles  are  tangent  to  a  given  line  at  a  given  point ; 
and  a  second  set  of  circles  are  tangent  to  a  second  line  at  the  same 
point.  What  is  the  locus  of  the  points  of  intersection  of  equal  circles 
of  the  two  families?    Prove  that  your  answer  is  correct. 

4.  Three  unequal  circles  are  so  situated  that  each  of  them  is  ex- 
ternally tangent  to  the  other  two.  At  the  points  of  contact  tangents 
are  drawn.    Prove  that  these  three  tangents  meet  in  a  point. 

191 


192  Plane  Geometry. 


*  5.   A  crescent-shaped  region  is  bounded  by  a  semicircumference 
of  radius  a,  and  another  circular  arc  whose  center  lies  on  the  semi- 


\ 

\ 
I 

I 


circumference  produced.     Find  the  area  and  the  perimeter  of  the 
region. 

6.  Prove  that  through  any  given  straight  line  a  plane  can  be  passed 
perpendicular  to  any  given  plane. 

Is  it  ever  possible  to  pass  more  than  one  such  plane  through  the 
line  ?  If  so,  when  ?  Can  a  plane  always  be  passed  through  a  given 
straight  line  perpendicular  to  a  given  line  ? 

7.  What  is  meant  by  the  pole  of  a  circle  on  a  sphere  ? 

Prove  that  all  the  points  on  the  circumference  of  a  circle  on  a  sphere 
are  equally  distant  from  either  of  its  poles. 
Define  the  term  "  polar  triangles." 

8.  AB,  AC,  AD  are  three  edges  of  a  cube  which  meet  in  the  vertex 
A .  A  plane  is  passed  through  the  middle  points  of  these  edges.  If 
the  cube  contains  8  cubic  feet,  find  the  volume  of  the  corner  cut  off  by 
the  plane,  and  the  length  of  the  perpendicular  dropped  from  the  center 
of  the  cube  on  the  plane. 

HARVARD  — 1900. 

1.  Prove  that  in  an  isosceles  triangle  the  angles  opposite  the  equal 
sides  are  equal. 

Given  two  lines  AB  and  BC.  Show  how  to  draw  through  a  point 
D  of  AB  a  third  line  making  with  BC  the  same  angle  which  AB 
makes  with  BC. 

2.  Prove  that  an  angle  formed  by  a  tangent  and  a  chord  is  measured 
by  one  half  the  intercepted  arc. 


Entrance  Examination   Papers.  193 

What  is  meant  here  by  the  statement  that  an  angle  is  "  measured 
by  "  half  an  arc  V 

A  certain  diameter  of  a  circle  makes  with  a  tangent  an  angle  of  70°. 
Find  the  angles  which  the  tangent  makes  with  the  chords  which  join 
the  point  of  contact  with  the  extremities  of  the  diameter. 

3.  Through  a  point  A  on  the  circumference  of  a  circle,  chords  are 
drawn.  On  each  one  of  these  chords  a  point  is  taken  one  third  of  the 
distance  from  A  to  the  other  end  of  the  chord.  Find  the  locus  of 
these  points,  and  prove  that  your  answer  is  correct. 

4.  Two  circles  intersect  at  the  points  A  and  B.  Prove  that  they 
make  equal  angles  with  one  another  at  A  and  at  B. 

[N.  B.  by  the  angle  between  two  circles  at  their  point  of  intersec- 
tion is  meant  the  angle  between  their  tangents  at  this  point.] 

An  area  (the  upper  area  in  the  figure)  is  bounded  by  three  arcs  of 
circles  which,  if  produced,  would  meet  in  a  point.     Prove  that  the 


sum  of  the  angles  between  the  circles  at  the  vertices  of  this  area  is 
equal  to  two  right  angles. 

5.  Prove  that  the  circumference  of  a  circle  is  the  limit  which  the 
perimeters  of  regular  inscribed  and  circumscribed  polygons  approach 
when  the  number  of  their  sides  is  increased  indefinitely. 

Find  the  perimeter  of  a  regular  hexagon  inscribed  in  a  circle  of 
radius  a ;  and  also  of  a  regular  hexagon  circumscribed  about  the 
same  circle.  What  can  you  infer  from  these  results  concerning  the 
value  of  TT  ? 

6.  A  square,  each  of  whose  sides  is  5  inches  long,  has  its  corners 
cut  off  in  such  a  way  as  to  make  it  into  a  regular  octagon.  Find  the 
area  and  the  perimeter  of  the  octagon. 


194  Plane  Geometry, 


HARVARD  — 1901. 

1.  Prove  that,  if  two  angles  of  a  triangle  are  equal,  the  triangle 
is  isosceles. 

The  triangle  ABC  has  a  right  angle  at  A.  Through  A  a  line  is 
drawn,  forming  with  AB  s^n  angle  equal  to  B,  and  cutting  BC  in  D. 
Prove  that  the  two  triangles  ABD  and  ADC  are  isosceles,  and  that  D 
is  the  middle  point  oi  BC. 

*2.  Prove  that,  when  two  tangents  to  the  same  circle  intersect 
each  other,  the  distances  from  their  point  of  intersection  to  their 
points  of  contact  are  equal. 

Two  circles  are  tangent  externally  at  J  ;  jBC  is  a  common  tangent, 
B  and  C  being  the  points  of  contact.  Prove  that  the  angle  BA  C  is  a 
right  angle. 

Hint:  Draw  the  common  tangent  at  ^. 

3.  Given  a  fixed  point  D  within  a  triangle  ABC.  Choose  any 
point,  E,  on  the  perimeter  of  the  triangle,  draw  DE,  and  let  P  be 
the  middle  point  of  DE.  Find  the  locus  of  P  when  E  traces  out  the 
whole  perimeter  of  the  triangle.  Describe  the  position  of  the  locus 
exactly,  and  prove  the  correctness  of  your  answer. 

4.  Two  parallel  tangents  are  drawn  to  a  circle,  and  a  third  tan- 
gent is  drawn  intersecting  these  two  in  A  and  B,  and  tangent  to  the 
circle  at  C.  Prove  that  the  product  of  the  segments  A  C  and  CB  is 
equal  to  the  square  of  the  radius  of  the  circle. 

*  5.  Given  a  straight  line  and  two  points,  one  of  which  lies  on  the 
line.  Show  how  to  construct  the  center  of  a  circle  passing  through 
the  two  points  and  tangent  to  the  line  at  the  given  point  of  the  line. 
Is  there  any  case  in  which  this  construction  is  impossible  ? 


*  6.   A  piece  is  cut  out  of  an  equilateral  triangle  by  means  of  an 
arc  of  a  circle  tangent  to  two  sides.     The  side  of  the  triangle  is  7 


Entrance  Examination  Papers.  195 

inches  and  the  radius  of  the  circle  1  inch.     Compute  to  two  decimal 
places  the  perimeter  and  the  area  of  the  figure  which  is  left. 

7.  Prove  that,  if  two  planes  are  perpendicular  to  each  other,  a 
straight  line  drawn  in  one  of  them  perpendicular  to  their  intersection 
is  perpendicular  to  the  other. 

Is  it  true  that  planes  perpendicular  to  the  same  straight  line  are 
parallel  to  each  other  ?     (Illustrate  by  a  figure.) 

Is  it  true  that  planes  perpendicular  to  the  same  plane  are  parallel 
to  each  other?     (Illustrate  by  a  figure.) 

8.  Prove  that  the  lateral  area  of  a  regular  pyramid  is  equal  to  the 
product  of  the  perimeter  of  its  base  and  half  its  slant  height. 

Is  this  theorem  true  for  an  irregular  pyramid?     (Give  the  reason.) 

9.  A  sphere  of  radius  13  inches  has  a  cylindrical  hole  bored  through 
it,  the  axis  of  the  cylinder  passing  through  the  center  of  the  sphere 
and  the  radius  of  the  cylinder  being  5  inches.  Find  the  entire  sur- 
face and  the  volume  of  the  solid  which  is  left. 


MASSACHUSETTS  INSTITUTE  OF  TECHNOLOGY  — 1900. 

1.  From  a  given  point  without  a  straight  line,  only  one  perpen- 
dicular can  be  drawn  to  the  line. 

2.  If  two  triangles  have  two  sides  of  one  equal  respectively  to  two 
sides  of  the  other,  but  the  third  side  of  the  first  greater  than  the  third 
side  of  the  second,  the  included  angle  of  the  first  is  greater  than  the 
included  angle  of  the  second. 

3.  The  medians  of  a  triangle  meet  in  a  point. 

4.  A  straight  line  can  intersect  a  circumference  in  not  more  than 
two  points. 

5.  In  the  same  circle,  or  in  equal  circles,  two  central  angles  are  in 
the  same  ratio  as  their  intercepted  arcs. 

6.  Two  triangles  are  similar  when  their  sides  are  parallel  each  to 
^each,  or  perpendicular  each  to  each. 

7.  If  the  triangle  ABC  has  the  side  AB  fixed,  and  the  side  AC 
twice  the  side  BC,  the  vertex  C  lies  on  a  circle  which  divides  the  line 
AB  externally  and  internally  in  the  same  ratio. 

8.  Compute  the  difference  in  area  between  a  circle  of  radius  3  and 
an  inscribed  regular  hexagon. 


196  Plane  Geometry. 


MASSACHUSETTS  INSTITUTE  OF   TECHNOLOGY— 1901. 

1.  The  perpendiculars  from  the  vertices  of  a  triangle  to  the  oppo- 
site sides  meet  in  a  common  point. 

2.  A  circle  is  inscribed  in  a  triangle  ABC.  If  D,  E,  and  F  are 
the  points  of  contact  of  the  sides  BC,  CA,  and  A  B,  respectively,  prove 
that  the  angle  EDF  is  one  half  the  supplement  of  the  angle  EA  F. 

3.  Two  circles  whose  diameters  are  8  and  2  inches,  respectively, 
touch  each  other  externally.  What  is  the  length  of  their  common 
tangent  ? 

4.  Define  Similar  Polygons.  Show  that  the  perimeters  of  two 
similar  polygons  are  in  the  same  ratio  as  any  two  homologous  sides. 

5.  In  equal  circles  or  in  the  same  circle  the  greater  of  two  chords 
subtends  the  greater  arc. 

6.  Show  how  to  construct  a  triangle  equivalent  to  a  given  polygon. 

7.  If  two  triangles  have  two  sides  of  the  one  equal,  respectively,  to 
two  sides  of  the  other,  but  the  third  side  of  the  first  greater  than  the 
third  side  of  the  second,  the  included  angle  of  the  first  is  greater  than 
the  included  angle  of  the  second. 

8.  If  the  radius  of  a  circle  is  6,  what  is  the  area  of  a  segment 
whose  arc  is  60°  ? 

BROWN  UNIVERSITY— 1900. 

1.  In  the  same  or  in  equal  circles  the  greater  of  two  unequal  chords 
is  nearer  the  center. 

2.  The  areas  of  two  rectangles  having  equal  altitudes  are  to  each 
other  as  their  bases. 

3.  A  line  drawn  from  a  vertex  of  a  triangle  to  the  middle  point  of 
the  opposite  side  is  less  than  half  the  sum  of  the  adjacent  sides. 

4.  Two  regular  polygons  of  the  same  number  of  sides  are  similar. 

5.  Find  the  side  of  a  square  inscribed  in  a  circle  whose  area  is  98  tt. 

BROWN  UNIVERSITY  — 1901. 

1.  Through  any  point  not  in  a  line  a  perpendicular  can  be  drawn 
to  the  line. 

2.  The  bisectors  of  the  angles  of  a  triangle  meet  in  a  point. 


Entrance  Examination  Papers.  197 

3.  Two  triangles  having  an  angle  of  one  equal  to  an  angle  of  the 
other  are  to  each  other  as  the  products  of  the  sides  including  the 
equal  angles. 

4.  A  regular  triangle  is  circumscribed  about  a  circle  whose  radius 
is  7.  Find  the  area  of  that  portion  of  the  triangle  which  is  outside  of 
the  circle. 

YALE  — 1899. 

1.  Two  triangles  are  similar  if  their  homologous  sides  are  propor- 
tional 

2.  Define  the  limit  of  a  variable.  Prove  that  if  two  variables  are 
always  equal  their  limits  are  equal.  Prove  that  the  area  of  a  circle  is 
equal  to  one  half  the  product  of  its  radius  and  circumference. 

3.  (a)  In  any  quadrilateral  if  a  line  be  drawn  through  the  middle 
points  of  two  adjacent  sides,  and  a  second  line  through  the  middle 
points  of  the  other  two  sides,  these  lines  will  be  parallel. 

(b)  If  the  middle  points  of  the  opposite  sides  of  a  quadrilateral  be 
joined,  the  lines  so  drawn  will  bisect  each  other. 

4.  (a)  If  two  circles  are  tangent  internally  and  if  the  radius  of  the 
one  be  the  diameter  of  the  other,  a  chord  of  the  larger  drawn  through 
the  point  of  tangency  is  bisected  by  the  smaller. 

(b)  What  example  of  a  locus  is  found  in  the  previous  figure  ? 


YALE  — 1900. 

[Time,  One  hour.] 

In  Question  1  the  work  will  not  be  accepted  unless  the  constructions  are 
made  accurately  with  ruler  and  compass. 

1.  Divide  a  line  externally  in  extreme  and  mean  ratio  and  prove 
the  construction. 

What  use  is  made  of  this  construction  in  Geometry  ? 

2.  (a)  From  a  point  A  on  the  circumference  of  a  circle  two  equal 
chords,  AB  and  AC,  are  drawn.  Prove  that  they  make  equal  angles 
with  the  diameter  through  A. 

(6)  If  each  of  these  angles  is  30°,  prove  ihat  the  points  A,  B,  and  C 
trisect  the  circumference. 


198  Plane  Geometry. 


3.  In  two  similar  triangles  the  bases  and  altitudes  are  proportional, 
and  in  two  equivalent  triangles  the  bases  and  altitudes  are  inversely 
proportional. 

4.  (a)  Prove  that  the  ratio  of  the  circumference  of  a  circle  to  its 
diameter  is  the  same  for  all  circles.  What  is  the  approximate  value 
of  this  ratio  ? 

(6)  Explain  briefly  (without  proofs)  the  method  of  determining 
this  value. 

YALE  — 1901. 

1.  State  and  prove  a  theorem  relating  to  the  squares  of  the  sides  of 
an  obtuse-angled  triangle. 

2.  Construct  accurately  with  ruler  and  compass  two  tangents  to  a 
circle  which  shall  include  between  their  points  of  contact  an  arc  of 
120°  of  the  circumference,  and  write  the  value  of  the  angA  included 
between  them. 

The  proof  may  he  omitted  if  the  method  is  made  clear  by  the  construction 
of  the  fgure. 

3.  Define  locus  of  a  point. 

Prove  that  the  bisector  of  an  angle  v4  of  a  triangle  and  the  bisectors 
of  its  exterior  angles  B  and  C  meet  in  a  point. 

4.  An  isosceles  triangle  with  vertex  A  is  inscribed  in  a  circle,  and 
through  A  a  line  is  drawn  cutting  the  side  BC  at  E  and  the  circle  at 
i).     Prove  that  AB  is  a  mean  proportional  between  AD  and  AE. 

5.  Quote  a  theorem,  a  definition,  and  an  axiom  used  in  proving 
that  the  area  of  a  triangle  is  equal  to  one  half  the  product  of  its  base 
and  altitude.     Write  nothing  else. 

SHEFFIELD   SCIENTIFIC   SCHOOL  — 1899. 

[Note.  —  State  at  the  head  of  your  paper  what  text-book  you  have  studied 
on  the  subject  and  to  what  extent.] 

1.  (a)  The  sum  of  the  angles  of  a  triangle  equals  two  right  angles. 
(6)  State  and  prove  the  theorem  on  the  sum  of  the  angles  of  any 

polygon. 

2.  If  two  circumferences  intersect,  the  straight  line  joining  their 
centers  bisects  their  common  chord  at  right  angles. 

State  the  corresponding  theorem  when  the  circumferences  are  tan- 
gent to  each  other. 


Entrance  Examination  Papers.  199 

3.  A  straight  line  parallel  to  one  side  of  a  triangle  divides  the 
other  two  sides  proportionally. 

4.  When  is  a  straight  line  divided  in  extreme  and  mean  ratio  ? 
Give  and  prove  the  construction  for  dividing  a  given  straight  line 

in  extreme  and  mean  ratio. 

To  the  construction  of  what  regular  polygon  does  this  construction 
apply? 

5.  What  is  the  meaning  of  tt  in  geometry  ? 

Find  the  length  of  the  side  of  an  equilateral  triangle  whose  area 
equals  that  of  a  circle  of  radius  R. 

SHEFFIELD  SCIENTIFIC   SCHOOL  — 1900. 

[Note.  —  State  at  the  head  of  your  paper  what  text-book  you  have  studied 
on  the  subject  and  to  what  extent.] 

1.  The  three  perpendiculars  erected  at  the  middle  points  of  the 
sides  of  a  triangle  meet  in  a  common  point. 

2.  State  and  prove  the  theorems  regarding  the  measurement  of  the 
angle  between  (a)  two  chords  of  a  circle ;  (b)  two  secants  of  a  circle. 

3.  AVhen  are  two  magnitudes  commensurable?  when  incommen- 
surable ? 

Prove  that  the  areas  of  two  rectangles  with  equal  bases  are  in  the 
same  ratio  as  the  altitudes,  both  when  the  latter  are  commensurable 
and  incommensurable. 

4.  The  areas  of  two  triangles  having  an  angle  of  the  one  equal  to 
an  angle  of  the  other  are  to  each  other  as  the  products  of  the  sides 
including  the  equal  angles. 

5.  Given  a  square  the  length  of  whose  side  is  6  units,  construct  a 
rectangle  with  altitude  2  units  and  equivalent  to  the  square. 

6.  What  is  the  meaning  of  tt  in  geometi'y?  State  and  prove  the 
theorem  on  the  area  of  a  circle. 

UNIVERSITY  OF  PENNSYLVANIA  — 1901. 
[Time,  Two  hours.] 

1.  Define :  A  curve,  parallel  straight  lines,  polygon  inscribed  in  a 
circle,  similar  figures,  harmonic  division  of  a  straight  line. 

2.  From  a  point  without  a  straight  line,  one  perpendicular  can  be 
drawn  to  that  line,  and  but  one. 


200  Plane  Geometry. 

3.  In  a  trapezoid  the  straight  line  joining  the  middle  points  of  the 
non-parallel  sides  is  parallel  to  the  bases,  and  is  equal  to  one  half 
their  sum. 

4.  An  angle  formed  by  two  secants  intersecting  without  the  cir- 
cumference is  measured  by  one  half  the  difference  of  the  intercepted 
arcs. 

5.  In  any  triangle,  the  sum  of  the  squares  of  the  two  sides  is  equal 
to  twice  the  square  of  half  the  base  increased  by  twice  the  square  of 
the  medial  line. 

6.  Triangles  on  the  same  or  equal  bases  and  between  the  same 
parallels  are  equivalent. 

7.  A  circle  may  be  circumscribed  about  any  regular  polygon ;  and 
a  circle  may  also  be  inscribed  in  it. 

8.  Problem  :  To  find  two  straight  lines  in  the  ratio  of  the  areas  of 
two  given  polygons. 

DARTMOUTH  — 1900. 

1.  Prove  that  the  three  perpendicular  bisectors  of  the  sides  of  any 
triangle  meet  in  a  common  point. 

2.  Circumscribe  a  circle  about  a  given  triangle. 

3.  Construct  a  circle  having  its  center  in  a  given  line,  and  passing 
through  two  given  points. 

4.  Prove,  algebraically,  that  the  square  of  the  side  opposite  the 
obtuse  angle  of  an  obtuse-angled  triangle  is  equal  to  the  sum  of  the 
squares  of  the  other  sides  plus  the  product  of  one  of  these  sides  and 
the  projection  of  the  other  side  upon  it. 

5.  If  the  side  of  an  equilateral  triangle  is  a,  find  its  area. 

6.  Prove  that  the  area  of  a  circle  is  equal  to  half  the  product  of  its 
radius  and  circumference. 

7.  The  radius  of  a  circle  is  2 ;  find  the  area  of  the  regular  inscribed 
dodecagon. 

8.  The  radius  of  a  circle  is  R ;  what  is  the  radius  of  a  concentric 
circle  which  divides  it  into  two  equivalent  parts  ? 


Entrance  Examination  Papers,         201 

DARTMOUTH  — 1901. 

1.  Prove  that  the  diagonals  of  a  parallelogram  bisect  each  other. 
When  are  they  equal  ? 

2.  How  many  degrees  in  one  angle  of  a  regular  decagon?  of  a 
regular  dodecagon?  What  is  the  largest  number  of  degrees  possible 
in  one  angle  of  a  regular  polygon? 

3.  One  angle  between  two  chords  intersecting  within  a  circle  is  50° ; 
its  intercepted  arc  is  10°.  How  many  degrees  in  the  arc  intercepted 
by  its  vertical  angle  ? 

4.  Prove  that  the  bisector  of  an  angle  of  a  triangle  divides  the 
opposite  side  into  segments  proportional  to  the  sides  of  the  angle. 

5.  The  diagonals  of  a  rhombus  are  9  and  12.  Find  its  perimeter 
and  area. 

6.  In  a  circle  whose  radius  is  R,  show  that  the  area  of  the  inscribed 
square  is  2  R^,  and  of  the  inscribed  regular  dodecagon  is  3  R^. 

7.  Draw  the  tangents  to  a  circle  from  a  point  outside  the  circle, 
and  prove  that  they  are  equal. 

PRINCETON  — 1901. 
Geometry  (a). 

1.  Two  triangles  are  equal  when  a  side  and  two  adjacent  angles  of 
the  one  are  equal  respectively  to  a  side  and  two  adjacent  angles  of  the 
other. 

2.  The  sum  of  the  interior  angles  of  a  polygon  is  equal  to  two  right 
angles,  taken  as  many  times  less  two  as  the  figure  has  sides. 

3.  The  tangents  to  a  circle  from  an  exterior  point  are  equal. 

4.  If  any  chord  is  drawn  through  a  fixed  point  within  a  circle,  the 
product  of  its  segments  is  a  constant  for  all  positions  of  the  chord. 

5.  Through  any  point  P  two  lines  are  drawn  ;  on  one  of  these,  two 
points  D  and  R  are  taken  on  opposite  sides  of  P,  and  on  the  other 
line  any  point  C  is  chosen ;  then  a  circle  is  described  through  the 
points  C,  Z),  and  P,  and  also  another  circle  through  the  point  R  and 
tangent  to  the  first  circle  at  the  point  P ;  the  line  CP  is  produced  to 
meet  the  second  circle  at  the  point  K.  Prove  the  angle  RKP  is  equal 
to  the  angle  PCD. 


202  Plane  Geometry. 

Geometry  (&).    Mensuration. 

Use  logarithms  in  eacli  exercise. 

1.  In  a  right  triangle  one  side  is  5.06  meters  and  the  hypothenuse  is 
12.4  meters ;  what  is  the  area  of  the  triangle  ? 

2.  The  areas  of  two  similar  triangles  are  3.42  square  meters  and 
8.04  square  meters,  and  the  altitude  of  the  smaller  is  1.25  meters; 
what  is  the  altitude  of  the  other  triangle  ? 

3.  A  quadrilateral  of  perimeter  36.4  meters  is  circumscribed  about 
a  circle  of  radius  3.62  meters;  what  is  the  area  of  the  portion  of  the 
quadrilateral  without  the  circle  ? 

4.  Find  the  area  of  the  ring  enclosed  between  two  concentric  circles 
of  radii  4.32  meters  and  3.41  meters. 

PRINCETON  SCHOOL  OF  SCIENCE  — 1901. 

1.  Prove  that  the  perpendicular  bisectors  of  the  sides  of  a  triangle 
meet  in  a  point,  and  that  this  point  is  equidistant  from  the  vertices  of 
the  triangle. 

2.  Define  when  a  plane  figure  is  symmetrical  with  respect  to  a  line ; 
also  with  respect  to  a  point. 

Prove  that  if  a  figure  is  symmetrical  with  respect  to  two  lines  per- 
pendicular to  each  other,  it  is  also  symmetrical  with  respect  to  their 
intersection. 

3.  Show  how  to  construct  geometrically  a  fourth  proportional  to 
three  given  straight  lines ;  also  how  to  construct  a  triangle  when  two 
sides  and  the  included  angle  are  given. 

4.  In  any  plane  triangle  state  and  prove  what  the  square  of  the  side 
opposite  an  acute  angle  is  equal  to. 

5.  Prove  what  the  area  of  a  triangle  is  equal  to ;  also  the  area  of  a 
trapezoid.     Define  a  trapezoid. 

6.  Show  and  prove  how  to  construct  a  square  equivalent  to  a  given 
parallelogram  ;  also  a  square  equivalent  to  a  given  triangle. 

7.  Define  a  regular  polygon.  Prove  that  the  perimeters  of  two  reg- 
ular polygons  of  the  same  number  of  sides  are  to  each  other  as  the 
radii  of  their  circumscribed  circles ;  and  that  their  areas  are  as  the 
squares  of  these  radii. 


Entrance  Examination  Papers.         203 


WORCESTER  POLYTECHNIC   INSTITUTE  — 1901. 

1.  Demonstrate :  If  from  a  fixed  point  without  a  circle  a  secant 
be  drawn,  the  product  of  the  secant  and  its  external  segment  is  con- 
stant, in  whatever  direction  the  secant  is  drawn. 

2.  Divide  a  given  straight  line  into  parts  proportional  to  any 
number  of  given  lines.     Give  proof. 

3.  (a)  When  is  a  line  said  to  be  divided  into  extreme  and  mean 
ratio  ?  (b)  What  is  a  rhombus  ?  (c)  A  regular  decagon  ?  (</)  The 
median  of  a  triangle?     (e)  An  oblique  angle? 

4.  Find  the  angle  at  the  center  of  a  circle  subtended  by  an  arc 
6  feet  5  inches  long,  if  the  radius  of  the  circle  is  8  feet  2  inches. 

5.  Demonstrate :  If  two  circles  intersect  and  a  secant  is  drawn 
through  each  point  of  intersection,  the  chords  which  join  the  ex- 
tremities of  the  secants  are  parallel. 

[Note. —No  credit  will  be  given  for  statements  made  in  the  course  of 
a  demonstration,  unless  the  proper  reasons  therefor  are  also  given.] 


CORNELL  — 1901. 

1.  Define  :  A  point,  a  straight  line,  a  plane,  two  parallel  straight 
lines,  a  plane  angle,  a  rectangle,  a  regular  polygon. 

2.  If  two  sides  of  a  triangle  be  unequal,  the  opposite  angles  are 
unequal,  and  the  greater  angle  lies  opposite  the  longer  side ;  and 
conversely. 

3.  At  a  given  point  to  construct  an  angle  equal  to  a  given  angle. 

4.  To  inscribe  a  circle  in  a  given  triangle,  and  to  escribe  three 
circles  to  it. 

5.  Three  or  more  parallels  will  cut  any  two  straight  lines,  not 
parallel  to  them,  in  proportional  segments. 

6.  To  construct  a  polygon  that  shall  be  similar  to  one  given 
polygon  and  equal  in  area  to  another. 

7.  Two  regular  polygons  of  the  same  number  of  sides  are  similar. 

8.  If  the  radius  of  a  circle  be  8  feet,  find  the  area  of  a  segment  cut 
off  by  the  side  of  an  inscribed  equilateral  triangle. 


204  Plane  Geometry, 


AMHERST  — 1901. 

1.  The  line  joining  the  middle  points  of  the  non-parallel  sides  of  a 
trapezoid  is  parallel  to  the  bases,  and  equal  to  one  half  their  sum. 

2.  The  angle  between  two  secants,  intersecting  without  the  cir- 
cumference, is  measured  by  one  half  the  difference  of  the  intercepted 
arcs. 

3.  Two  polygons  are  similar  when  they  are  composed  of  the  same 
number  of  triangles,  similar  each  to  each,  and  similarly  placed. 

4.  To  construct  a  square  equivalent  to  twice  a  given  square. 

5.  The  perimeters  of  two  regular  polygons  of  the  same  number  of 
sides  are  to  each  other  as  their  radii,  or  as  their  apothems. 

6.  The  area  of  an  equilateral  triangle  is  9  VH.     Find  its  side. 


WELLESLEY  COLLEGE  — 1901. 

1.  State  three  cases  in  which  triangles  are  similar.  Prove  one  of 
them. 

2.  A  circle  can  be  circumscribed  about  any  regular  polygon. 

3.  If  one  of  the  equal  sides  of  an  isosceles  triangle  be  the  diameter 
of  a  circle,  the  circumference  will  bisect  the  base. 

4.  Solve  one : 

(a)  The  non-parallel  sides  of  a  trapezoid  are  13  each ;  the  parallel 
sides  are  50  and  26.     Find  its  area. 

(b)  Find  the  area  of  an  equilateral  triangle  whose  side  is  12. 

5.  If  two  triangles  have  two  sides  of  the  one  equal  respectively  to 
two  sides  of  the  other  but  the  included  angles  unequal,  what  conclu- 
sion can  be  drawn  ?    Prove. 

6.  Construct  a  mean  proportional  to  two  given  straight  lines,  giving 
reasons  for  work. 

7.  If  two  chords  intersect  within  a  circle,  what  is  the  measure  of 
the  angle  formed  ?    Prove. 


Entrance  Examination  Papers.         205 

SMITH  COLLEGE  — 1901. 

1.  The  sum  of  two  lines  drawn  from  a  point  to  the  extremities  of 
a  straight  line  is  greater  than  the  sum  of  two  other  lines  similarly 
drawn  but  included  by  them. 

2.  Through  a  given  point  without  a  circle  construct  a  tangent. 

3.  (Original.)  The  bisectors  of  the  angles  of  a  parallelogram  form 
a  rectangle. 

4.  Given  a  pentagon,  construct  an  equivalent  triangle. 

5.  Given  a  radius,  and  a  side  of  a  regular  inscribed  polygon,  find 
the  side  of  a  regular  polygon  of  double  the  number  of  sides. 

6.  (Original.)  The  sum  of  the  perpendiculars  from  any  point 
within  a  convex  equilateral  polygon  upon  the  sides  is  constant. 

VASSAR  COLLEGE  — 1901. 

1.  If  the  angle  at  the  vertex  of  an  isosceles  triangle  is  50°,  find  the 
exterior  angle  formed  by  producing  the  base. 

2.  Find  the  value  of  each  angle  of  an  equiangular  decagon. 

3.  Two  tangents  drawn  to  a  circle  from  an  exterior  point  form  an 
angle  of  60°.  How  many  degrees  are  there  in  the  arc  contained 
between  their  points  of  contact  ? 

4.  Prove  that  the  opposite  angles  of  an  inscribed  quadrilateral  are 
supplementary. 

5.  Prove  that  triangles  which  have  an  angle  in  each  equal  are  to 
each  other  as  the  rectangles  of  the  including  sides. 

6.  Define  and  illustrate  by  figures  supplementary  and  complemen- 
tary angles,  similar'polygons.  Give  all  the  conditions  of  the  equality 
of  the  triangles. 

7.  How  many  tiles  9  inches  long  and  4  inches  wide  will  be  required 
to  pave  a  path  8  feet  wide  surrounding  a  rectangular  court  60  feet 
long  and  36  feet  wide? 

8.  Two  tangents  drawn  from  a  point  exterior  to  a  circle  whose 
radius  is  1.3  inches  intercept  upon  the  circumference  an  arc  whose 
length  is  0.57  inches.     Required  the  angle  formed  by  the  two  tangents. 


2o6  Plane  Geometry. 

UNIVERSITY  OF  CHICAGO  — 1901. 

[Time  Allowed,  One  Hour  and  Thirty  Minutes.] 

[When  required,  give  all  reasons  in  full,  and  work  out  proofs  and 
problems  in  detail.] 

1.  The  square  on  the  side  of  a  triangle  which  lies  opposite  a  right 
angle  is  equivalent  to  the  sum  of  the  squares  of  the  other  two  sides. 

(1)  Give  a  proof  of  this  theorem  different  from  the  one  given  in 
your  text-book. 

(2)  State  how  this  theorem  should  be  modified  when  the  given 
side  lies  opposite  (a)  an  acute  angle,  (&)  an  obtuse  angle. 

(3)  If  unable  to  comply  with  the  condition  (1),  give  a  proof  for 
either  (a)  or  (h)  in  (2). 

2.  Prove  that  the  area  bounded  by  the  circumferences  of  two  con- 
centric circles  of  unequal  radii  is  equivalent  to  the  area  of  a  circle 
whose  diameter  is  that  chord  of  the  outer  circumference  which  is 
tangent  to  the  inner  circumference. 

3.  Two  tangents  to  a  circle  meet  at  an  angle  of  60°.  If  the  radius 
of  the  circle  is  21  inches,  find  the  length  of  the  arc  included  between 
the  points  of  tangency  of  the  two  tangents.     [Use  tt  =  3^.] 

4.  Prove  that  one  of  the  angles  formed  at  the  intersection  of  the 
two  bisectors  of  the  base  angles  of  an  isosceles  triangle  is  equal  to 
one  of  the  exterior  base  angles  of  the  triangle. 


MICHIGAN  SCHOOL  OF  MINES— 1901. 

1.  Two  triangles  are  equal  when .     State  and  prove. 

2.  The  sum  of  the  interior  angles  of  a  polygon  is  equal  to . 

State  and  prove. 

3.  In  the  same  or  equal  circles  two  angles  at  the  center  have  the 
same  ratio  as  the  arcs  which  they  subtend  (two  cases).     Prove. 

4.  The  square  on  the  hypothenuse  of  a  right-angled  triangle  is 
equivalent  to  the  sum  of  the  squares  on  the  other  two  sides.    Prove. 


Entrance  Examination  Papers.  207 

5.  A  line  drawn  from  the  vertex  of  a  triangle  bisecting  the  vertical 
angle  divides  the  base  into  segments  which  are  proportional  to  the 
adjacent  sides.    Prove. 

6.  The  area  of  a  circle  is  equal  to .     State  and  prove. 


WASHINGTON"   UNIVERSITY  — 1900. 

1.  Prove  that  two  angles  whose  sides  are  respectively  perpendicu- 
lar to  each  other  are  either  equal  or  supplementary. 

2.  Prove  that  an  exterior  angle  of  a  triangle  is  equal  to  the  sum 
of  the  two  opposite  interior  angles. 

3.  Prove  that  the  three  lines  drawn  from  the  vertices  of  the  angles 
of  a  triangle  to  the  middle  points  of  the  opposite  sides  meet  at  a  point. 

4.  If  we  have  the  proportion 

^  =  ^,  prove  that  we  also  have^^i^  ^m±n^ 
b      n  a  —  b     m  —  n 

Express  this  in  words,  as  a  theorem. 

5.  Prove  that  the  line  which  bisects  an  angle  of  a  triangle  divides 
the  opposite  side  into  segments  proportional  to  the  adjacent  sides. 

6.  Suppose  two  intersecting  straight  lines  cut  across  the  same 
circle.  Prove  that  the  angle  between  them  is  measured  by  one  half 
the  difference  of  the  two  intercepted  arcs.  [Draw  the  figure  so  that 
one  of  the  lines  passes  through  the  center  of  the  circle,  wkile  the 
other  intersects  it  without  the  circle.] 

7.  Prove  that  if  two  triangles  have  their  homologous  sides  propor- 
tional, their  homologous  angles  are  equal. 


UNIVERSITY  OF  CALIFORNIA  — 1901. 

1.  Prove  that  the  sum  of  the  angles  of  a  triangle  is  equal  to  two 
right  angles.  State  a  very  important  axiom  (postulate)  upon  which 
the  proof  of  this  proposition  depends. 

2.  Prove  that  the  angle  contained  by  the  bisectors  of  two  exterior 
angles  of  any  triangle  is  equal  to  half  the  sum  of  the  two  correspond- 
ing interior  angles. 


2o8  Plane  Geometry. 

3.  Construct  a  triangle  having  given  its  base,  one  of  its  sides,  and 
its  altitude. 

4.  Prove  that  the  areas  of  similar  triangles  are  to  one  another  as 
the  squares  described  on  their  homologous  sides.  Show  that  this 
proposition  is  also  true  of  similar  polygons. 

5.  Two  chords  intersect  within  a  circle ;  prove  that  the  rectangle 
contained  by  the  segments  of  one  of  the  chords  is  equal  to  the  rec- 
tangle contained  by  the  segments  of  the  other. 

6.  From  a  point  P,  outside  a  circle,  a  straight  line  is  drawn  cut- 
ting the  circumference  in  M  and  N  so  that  PN  =  kPM.  The  radius 
of  the  circle  is  r  and  the  distance  of  P  from  the  center  is  d.  Find 
PM  in  terms  of  k,  r,  and  d. 


ADVERTISEMENTS. 


ESSENTIALS  OF  ALGEBRA 

By  WEBSTER  WELLS,  S.B., 

Professor  of  Mathematics  in  the  Massachusetts  Institute  of  Technology. 


This  book  fully  meets  the  most  rigid  requirements  now  made 
in  secondary  schools.  Like  the  author's  other  Algebras,  it  has 
met  with  marked  success  and  is  in  extensive  use  in  schools  of 
the  highest  rank  in  all  parts  of  the  country. 

The  method  of  presenting  the  fundamental  topics  differs  at 
several  points  from  that  usually  followed.  It  is  simpler,  more 
logical  and  more  philosophical,  yet  by  reason  of  its  admirable 
grading  and  superior  clearness  The  Essentials  of  Algebra  is  not 
a  difficult  book. 

The  examples  and  problems  number  over  three  thousand  and 
are  very  carefully  graded.  They  are  especially  numerous  in  the 
important  chapters  on  Factoring,  Fractions,  and  Radicals.  All 
of  them  are  new,  not  one  being  a  duplicate  of  a  problem  in 
the  author's  Academic  Algebra. 

In  accurate  definitions,  clear  and  logical  demonstrations,  well 
selected  and  abundant  problems,  in  systematic  arrangement  and 
completeness,  this  Algebra  is  unequalled. 


Half  leather.    ) ^8 pages.     Introduction  price^$i.io. 


D.  C.  HEATH  &  CO.,  Publishers,  Boston,  New  York,  Chicago 


EXERCISE  BOOK  IN  ALGEBRA 

^Designed  for  supplementary  or  review  work  in  connection  with 
any  text-book  in  tAlgebra. 

By  MATTHEW  S.  McCURDY,  M.A., 

Instructor  in  Mathematics  in  the  Phillips  Academy,  Andover,  Mass. 


This  book  is  designed  to  furnish  a  collection  of  exercises  similar 
in  character  to  those  in  the  ordinary  text-books,  of  medium  grade 
as  to  difficulty,  and  selected  with  special  reference  to  giving  an 
opportunity  for  drill  upon  those  subjects  which  experience  has 
shown  to  be  difficult  for  students  to  master. 

Though  intended  primarily  to  be  supplementary  to  some  regu- 
lar text-book,  a  number  of  definitions  and  a  few  rules  have  been 
added,  in  the  hope  that  it  may  also  be  found  useful:  as  an  inde- 
pendent review  and  drill  book.     With  or  without  answers. 

Cloth.    Pages,  vi  -\-  220.    Introduction  price,  60  cents. 


ALGEBRA  LESSONS 

By  J.  H.  GILBERT. 

This  series  is  intended  for  supplementary  or  review  work,  and 
contains  three  numbers :  No.  i  —  To  Fractional  Equations, 
No.  2 — Through  Quadratic  Equations,  No.  3  — Higher  Algebra. 

Paper.     Tablet  form.    Price,  $1.44  per  do^en. 


REVIEW  AND  TEST  PROBLEMS 
IN   ALGEBRA 

By  S.  J.  PETERSON  AND  L.  F.  BALDWIN. 

The  problems  in  this  manual  are  original  —  none  have  been 
copied  from  any  other  author.     They  illustrate  points  of  special 
importance,  and  are  sufficiently  varied  and  difficult  for  written 
drills  for  those  preparing  for  college  entrance  examinations. 
Paper.    8 y  pages.    Introduction  price,  ^o  cents. 

D.  C.  HEATH  &  CO.,  Publishers,  Boston,  New  York,  Chicago 


WELLS'S  COMPLETE  TRIGONOMETRY 

(1900) 

In  this  new  Trigonometry  many  improvements  have  been  made, 
notably  in  the  proofs  of  several  of  the  functions,  in  the  general  demon- 
strations of  the  formulae,  in  the  solution  of  right  triangles  by  natural 
fimctions,  etc.  The  book  contains  an  unusually  large  number  of  ex- 
amples. These  have  been  selected  with  great  care,  and  most  of  them 
are  new.      The  Table  of  Contents  shows  its  scope  : 

CHAPTER   I.  — Trigonometric  Functions  of  Acute  Angles. 

II.  — Trigonometric  Functions  of  Angles  in  General. 

III.  —  General  Formulae. 

IV.  —  Miscellaneous  Theorems,  including  Circular  Measure  of  the  Angle  ; 

Inverse  Trigonometric  Functions ;  Line  Values  of  the  Six  Func- 
tions i  Limiting  values  of ■  and ■ 

X  X 

V.  —  Logarithms  (Properties  and  Application). 

VI.  — Solution  of  Right  Triangles  ;  Formulae  for  arcs  of  Right  Triangles. 
VII.  —  General    Properties  of  Triangles  j    Formulae  for    arcs  of  Oblique 

Triangles. 
VIII.  —  Solution  of  Oblique  Triangles. 
IX.  —  Geometrical  Principles. 
X.  —  Right  Spherical  Triangles  (  Solution) . 

XI.  —  Oblique  Spherical  Triangles  (General    Properties,   Napier's   Ana- 
logies, Solution).  * 
XII.  —  Applications,  Formulae,  Answers,  Use  of  Tables. 

Attention  is  particularly  invited  to  the  following  features  : 

1.     The  proofs  of  the  functions  of  120°,  135°,  150°,  etc. 

a.     The  proofs  of  the  functions  of  ( — A)  and  (90°  -|-  A)  in  terms  of  those  of  A. 

3.  The  expression  of  the  function  of  any  angle,  positive  or  negative,  as  a  function 

of  a  certain  acute  angle. 

4.  The  general  demonstration  of  the  formulae  tan*-  = and 

COSJf 

sin^A?  -\-  cos^a;  =  I. 

5.  Also  of  cotaf  =  — ; — ,  sec^A?  =   i   _L_  tanajf  and  csc^Xz=.  I  -f-cot'*- 

sinr 

6.  The  proofs  of  the  formulae  for  i\n[x  -\-y)   and  co&{^x -\- y  )  when  *•  and  y 

are  acute,  and  when  x  -{-  y  \s  acute  or  obtuse. 

7.  The  proofs  of  the  formulae,  tan  I  X    '  ~  ^^^'^  and  cot  i  at  =  -L+i2?f. 

2  sinAT  2  sinAf 

8.  The  solution  of  right  triangles  by  Natural  Functions. 

9.  The  solution  of  quadrantal  and  isosceles  spherical  triangles. 
10.     The  solution  of  oblique  spherical  triangles. 

The  results  have  been  worked  out  by  aid  of  the  author's  New 
Four  Place  Tables. 

Complete  Trigonometry^  Half  Leather^  ^5^  PP-  QO  cts.y  wit  A  Four  Place  TabIeSy$T.o8. 
Plane  Trigonometry ^  Chapters  I-FIIIy  100  pp.  docts.^  ivitb  Four  Place  Tables^  7S  cts. 


History. 


Allen's  History  Topics.  Covers  Ancient,  Modern,  and  American  history  and  gives  aa 
excellent  list  of  books  of  reference.     121  pages.     Paper,  25  cents. 

Allen's  Topical  Outline  of  English  History.  Including  references  for  literature.  Boards, 
25  cents ;  cloth,  40  cents. 

Boutwell's  The  Constitution  of  the  United  States  at  the  End  of  the  First  Century. 
Presents  the  Constitution  as  it  has  been  interpreted  by  decisions  of  the  United  States  Su- 
preme Court  from  1789  to  1889.     430  pages.     Buckram,  ;^2.so ;  law  sheep,  ^^3.50. 

Fisher's  Select  Bibliography  of  Ecclesiastical  History.  An  annotated  list  of  the  most 
essential  books  for  a  theological  student's  library.     15  cents. 

Flickinger's  Civil  Government:  as  Developed  in  the  States  and  the  United  States. 
An  historical  and  analytic  study  of  civil  institutions,  for  schools  and  colleges.  374  pages. 
Cloth,  $1.00, 

Hall's  Method  of  Teaching  History.  "  Its  excellence  and  helpfulness  ought  to  secure  it 
many  readers." — The  Nation.    405  pages.    ;j5i.5o. 

Pratt's  America's  Story  for  America's  Children.  A  series  of  history  readers  for  ele- 
mentary schools. 

I.    The  Beginner's  Book.    Cloth.    60  illustrations.    132  pages.    35  cents. 
II.    Discoverers  and  Explorers:   1000  to  1609.  Cloth.     152  pages.     52  illus.    40  cents. 
III.    The  Earlier  Colonies:   1601  to  1733.     Cloth.     160  pages.     Illus.    40  cents. 
IV.    The  Later  Colonies.    Cloth.     Illus.     160  pages.    40  cents. 
V.    The  Revolution  and  the  Republic.    Cloth.    Illus.     160  pages.    40  cents. 

Sheldon's  American  History.  Follows  the  "  seminary  "  or  laboratory  plan.  "  By  it  the 
pupil  is  not  robbed  of  the  right  to  do  his  own  thinking."     Half  leather.    $1.12. 

Teacher's  Manual  to  Sheldon's  American  History.    60  cents. 

Sheldon's  General  History.  For  high  schools  and  colleges.  The  only  general  history 
following  the  "  seminary  "  or  laboratory  plan.     Half  leather.    572  pages.    i>\.(x>. 

■Teacher's  Manual  to  Sheldon's  History.  Puts  into  the  instructor's  hand  the  key  to  the 
above  system.     172  pages.    85  cents. 

Sheldon's  Greek  and  Roman  History.  Contains  the  first  250  pages  of  the  General 
History,    ^i.oo. 

Sheldon-Barnes's  Studies  in  Historical  Method.  Suggestive  studies  for  teachers  and 
students.    Cloth.     160  pages.    90  cents. 

Shumway'S  A  Day  in  Ancient  Rome.  With  59  illustrations.  Should  find  a  place  as  a 
supplementary  reader  in  every  high-school  class  studying  Cicero,  Horace,  Tacitus,  etc. 
96  pages.    Paper,  30  cents ;  cloth,  75  cents. 

Thomas's  Elementary  History  of  the  United  States.  For  younger  grades.  Maps  and 
illustrations.     Cloth.     357  pages.    60  cents. 


Thomas's  History  of  the  United  States.  Revised  and  rewritten.  Edition  of  1901.  For 
schools,  academies,  and  the  genera)  reader.  A  narrative  history  with  copious  references 
to  sources  and  authorities.     Fully  illustrated.     592  pages.     Half  leather.    $1,00. 

English  History  Readers.     English  history  for  grammar  grades. 

Wilson's  Compendium  of  United  States  and  Contemporary  History.  For  schools  and 
the  general  reader.     114  pages.    40  cents. 

"Wilson's  The  State.  Elements  of  Historical  and  Practical  Politics.  A  book  on  the 
organization  and  functions  of  government.  Revised  edition,  largely  rewritten.  693 
pages.    $2.00. 

Sent  postpaid  on  receipt  of  price  by  the  publishers. 

D.  C.  HEATH  &  CO.,  Publishers,Boston,  NewYork,Chicago 


Heath's   English   Classics. 


Addison's  Sir  Roger  de  Coverley  Papers.  Edited  by  W.  H.  Hudson,  Professor  in  the 
Leland  Stanford  Junior  University.  Cloth.  232  pages.  Nine  full-page  illustrations  and 
two  maps.    35  cents. 

Burke's  Speech  on  Conciliation  with  America.  Edited  by  A.  J.  George,  Master  in 
the  Newton  (Mass.)  High  School.     Cloth.     119  pages,    accents. 

Carlyle'S  Essay  on  Bums.  Edited,  with  introduction  and  notes,  by  Andrew  J.  Georgr. 
Cloth.     159  pages.     Illustrated.     25  cents. 

Coleridge's  Rime  of  the  Ancient  Mariner.     Edited  by  Andrew  J.  George.    Cloth.    96 

pages.     Illustrated.     20  cents. 

Cooper's  Last  of  the  Mohicans.  Edited  by  J.  G.  Wight,  Principal  Girls'  High  School, 
New  York  City.     Cloth.     Maps  and  illustrations.     659  pages.     50  cents. 

DeQuincey'S  Flight  of  a  Tartar  Tribe.  Edited  by  G.  A.  Wauchope,  Professor  in  the 
University  of  South  Carolina.     Cloth.     112  pages.    25  cents. 

Dryden's  Palamon  and  Arcite.  Edited  by  William  H.  Crawshaw,  Professor  in  Colgate 
University.     Cloth.     158  pages.     Illustrated.     25  cents. 

George  Eliot's  Silas  Mamer.  Edited  by  G.  A.  Wauchope,  Professor  in  the  University  of 
South  Carolina.     Cloth.     288  pages.     Illustrated.     35  cents. 

Goldsmith's  Vicar  of  Wakefield.  With  introduction  and  notes  by  William  Henry 
Hudson.    Cloth.    300  pages.   Seventeen  full-page  illustrations  by  C.  E.  Brock.    50  cents. 

Macaulay'S  Essay  on  Milton.  Edited  by  Albert  Perry  Walker,  editor  of  Milton's 
"  Paradise  Lost,"  Master  in  the  English  High  School,  Boston.  Cloth.  146  pages. 
Illustrated.    25  cents. 

Macaulay'S  Essay  on  Addison.  Edited  by  Albert  Perry  Walker.  Cloth.  192  pages. 
Illustrated.     25  cents. 

Milton's  Paradise  Lost.    Books  I  and  II.     Edited  by  Albert  Perry  Walker.    Cloth. 

188  pages.     Illustrated.    25  cents. 

Milton's  Minor  Poems.  Edited  by  Albert  Perry  Walker.  Cloth.  190  pages.  Illus- 
trated.   25  cents. 

Pope's  Translation  of  the  Iliad.  Books  I,  VI,  XXII,  and  XXIV.  Edited  by  Paul 
Shorev,  Professor  in  the  University  of  Chicago.  Cloth.  174  pages.  Illustrated. 
25  cents. 

Scott's  Ivanhoe.  Edited  by  Porter  Lander  MacClintock,  Instructor  in  the  University 
of  Chicago.  Cloth.  556  pages.  Seventeen  full-page  illustrations  by  C.  E.  Brock. 
50  cents. 

Shakespeare '  S  Macbeth .  Edited  by  Edmund  K.  Chambers.  In  the  A  rden  Shakespeare 
series.     Cloth.     188  pages.     25  cents. 

Shakespeare's  Merchant  of  Venice.  Edited  by  H.  L.  Withers.  In  the  Arden  Shake- 
speare s&n&s.    Cloth.     178  pages.     25  cents. 

Tennyson's  Enoch  Arden  and  the  two  Locksley  Halls.  Edited  by  Calvin  S.  Brown, 
Professor  in  the  University  of  Colorado.     Cloth.     168  pages.     25  cents. 

Tennyson's  The  Princess.  With  introduction  and  notes  by  Andrew  J.  George,  Master 
in  the  Newton  (Mass.)  High  School.     Cloth.     148  pages.     Illustrated.     25  cents. 

Webster's  First  Bunker  Hill  Oration.  With  introduction  and  notes  by  Andrew  J.  George. 
Boards.    55  pages.    20  cents. 

See  also  our  lists  0/ books  in  English  Literature  and  Higher  English. 

D.  C.  HEATH  &  CO.,  Publishers,  Boston,  New  York,  Chicago 


Science. 


Ballard's  World  of  Matter.     A  guide  to  mineralogy  and  chemistry.    5i-oo« 

Benton's  Guide  to  General  Chemistry.     A  manualfor  the  laboratory.    35  cents. 

Boyer's  Laboratory  Manual  in  Biology.  An  elementary  guide  to  the  laboratory  study  of 
animals  and  plants.    80  cents. 

Boynton,  Morse  and  Watson's  Laboratory  Manual  in  Chemistry.    50  cents. 

Chute's  Physical  Laboratory  Manual.  A  well-balanced  course  in  laboratory  physics,  re- 
quiring inexpensive  apparatus.     Illustrated.    80  cents. 

Chute's  Practical  Physics.     For  high  scnools  and  colleges.    $1.12. 

Clark's  Methods  in  Microscopy.     Detailed  descriptions  of  successful  methods.    $i.6o. 

Coit's  Chemical  Arithmetic.     With  a  short  system  of  analysis.     50  cents. 

Colton's  Physiology  :  Experimental  and  Descriptive.  For  high  schools  and  colleges. 
Illustrated.    $1.12. 

Colton's  Physiology:  Briefer  Course.  For  earlier  years  in  high  schools.  Illustrated, 
go  cents. 

Colton's  Practical  Zoology.     Gives  careful  study  to  typical  animals.    60  cents. 

Grabfleld  and  Burns's  Chemical  Problems.     For  review  and  drill.     Paper.    25  cents. 

Hyatt's  Insecta.     A  practical  manual  for  students  and  teachers.     Illustrated.    ^1.25. 

Wewell's  Experimental  Chemistry.     A  modern  text-book  in  chemistry  for  high  schools 

and  colleges,    ^i.io. 
Orndorff 's  Laboratory  Manual.    Contains  directions  for  a  course  of  experiments  in  Organic 

Chemistry,  arranged  to  accompany  Remsen's  Chemistry.     Boards.     35  cents. 

Pepoon,  Mitchell  and  Maxwell's  Plant  Life.    A  laboratory  guide.    50  cents. 

Remsen's  Organic  Chemistry.  An  introduction  to  the  study  of  the  compounds  of  carbon. 
For  students  of  the  pure  science,  or  its  application  to  arts.    $1.20. 

Roberts's  Stereo-Chemistry.     Its  development  and  present  aspects,     i.oo. 

Sanford'S  Experimental  Psychology.     Parti.     Sensation  and  Perception.    $1.50. 

Shaler's  First  Book  in  Geology.     Cloth,  60  cents.    Boards,  45  cents. 

Shepard '  s  Inorganic  Chemistry.  Descriptive  and  qualitative  ;  experimental  and  inductive ; 
leads  the  student  to  observe  and  think.     For  high  schools  and  colleges.    ^1.12. 

Shepard 'S  Brief  er  Course  in  Chemistry,  with  chapter  on  Organic  Chemistry.  For  schools 
giving  a  half  year  or  less  to  the  subject,  and  schools  limited  in  laboratory  facilities.  80  cents. 

Shepard'S  Laboratory  Note-Book.  Blanks  for  experiments  ;  tables  for  the  reactions  of 
metallic  salts.     Can  be  used  with  any  chemistry.     Boards.    35  cents. 

Spalding's  Botany.     Practical  exercises  in  the  study  of  plants.    80  cents. 

Stevens's  Chemistry  Note-Book.     Laboratory  sheets  and  covers.     50  cents. 

Venable's  Short  History  of  Chemistry.     For  students  and  the  general  reader.    $1.00. 

Walter,  Whitney  and  Lucas's  Animal  Life.    A  laboratory  guide.    50  cents. 

Whiting's  Physical  Measurement.  I.  Density,  Heat,  Light,  and  Sound.  II.  Dynamics, 
Magnetism,  Electricity.  III.  Principles  and  Methods  of  Physical  Measurement,  Physi- 
cal Laws  and  Principles,  and  Tables.     Parts  I-IV,  in  one  volume,  $i.7S' 

Whiting's  Mathematical  and  Physical  Tables.    Paper.    50  cents. 

Williams's  Modern  Petrography.     Paper.    25  cents. 

For  elementary  ivorks  see  our  list  0/ 
books  in  Elementary  Science, 

D.C.  HEATH  &  CO.,  Publishers,  Boston,  NewYork,  Chicago 


Mathematics 


Barton's  Theory  of  Equations.    A  treatise  for  college  classes.    $1.50. 

Bowser's  Academic  Algebra.    For  secondary  schools.    $1.12. 

Bowser's  College  Algebra.     A  full  treatment  of  elementary  and  advanced  topics.    $1.50, 

Bowser»8  Plane  and  Solid  Geometry.    $1.25.    Plane,  bound  separately.    75  cts. 

Bowser's  Elements  of  Plane  and  Spherical  Trigonometry.    90  cts.;  with  tables,  $1.40. 

Bowser's  Treatise  on  Plane  and  Spherical  Trigonometry.  An  advanced  work  for  col- 
leges and  technical  schools.    $1.50. 

Bowser's  Five-Place  Logarithmic  Tables.    50  cts. 

Fine's  Number  System  in  Algebra.    Theoretical  and  historical.    $1.00. 

Gilbert's  Algebra  Lessons.  Three  numbers :  No.  i,  to  Fractional  Equations;  No.  2. 
through  Quadratic  Equations;  No.  3,  Higher  Algebra.     Each  number,  per  dozen,  $1.44. 

Hopkins's  Plane  Geometry.     Follows  the  inductive  method.     75  cts. 

Howland's  Elements  of  the  Conic  Sections.    75  cts. 

Lefevre'8  Number  and  its  Algebra.     Introductory  to  college  courses  in  algebra.     $1.25. 

Lyman's  Geometry  Exercises.     Supplementary  work  for  drill.     Per  dozen,  $1.60. 

McCurdy'8  Exercise  Book  in  Algebra.    A  thorough  drill  book.    60  cts. 

Miller's  Plane  and  Spherical  Trigonometry.  For  colleges  and  technical  cchools.  $1.15. 
With  six-place  tables,  $1.40. 

RiohOl'S  Analytic  Geometry.     A  treatise  for  college  courses.    $1.35. 

Nichols's  Calculus.     Differential  and  Integral.     ;f2.oo. 

Osborne's  Differential  and  Integral  Calculus.    $2.00. 

Peterson  and  Baldwin's  Problems  in  Algebra.    For  texts  and  reviews.    30  ct». 

Robbins's  Surveying  and  Navigation.    A  brief  and  practical  treatise.    50  cts. 

Schwatt's  Geometrical  Treatment  of  Curves.    $1.00. 

Waldo's  Descriptive  Geometry.  A  large  number  of  problems  systematically  arranged  and 
with  suggestions.     80  cts. 

Wells's  Academic  Arithmetic.    With  or  without  answers.    $1.00. 

Wells's  Essentials  of  Algebra.    For  secondary  schools.    $1.10. 

Wells's  Academic  Algebra.     With  or  without  answers.     $1.08. 

Wells's  New  Higher  Algebra.    For  schools  and  colleges.    $1.32. 

Wells's  Higher  Algebra.    $1.32. 

Wells's  University  Algebra.    Octavo.    $1.50. 

Wells's  College  Algebra.     $1.50.     Part  II,  beginning  with  quadratics.     $1.32. 

Wells's  Essentials  Of  Geometry.    (1899.)    $1.25.    Plane,  75  cts.    Solid,  75  cts. 

Wells's  Elements  of  Geometry.    Revised.    (1894.)    $1.25.    Plane,  75  cts.;  Solid,  75  cts. 

Wells's  New  Plane  and  Spherical  Trigonometry.  For  colleges  and  technical  schools. 
$1.00.    With  six  place  tables,  $1.25.    With  Robbins's  Surveying  and  Navigation,  $1.50. 

Wells's  Complete  Trigonometry.  Plane  and  Spherical.  90  cts  With  tables,  $1.08. 
Plane,  bound  separately,  75  cts. 

Wells's  New  Six-Place  Logarithmic  Tables.    60  cts. 

Wells's  Four-Place  Tables.    25  cts. 

For  Arithmetics  see  our  list  of  books  in  Elementary  Mathemaiiet. 

D.C.  HEATH  &  CO., Publishers, Boston,  New  York, Chicago 


Drawing    and    Manual    Training. 

Thompson's  New  Short  Course  in  Drawing.  A  practical,  well-balanced  sys- 
tem, based  on  correct  principles.  Can  be  taught  by  the  ordinary  teacher  and  learned  by 
the  ordinary  pupil.  Books  I-IV,  6x9  inches,  per  dozen,  ^1,20.  Books  V-VIII,  9x12 
inches,  per  dozen,  $1.75.  Manual  to  Books  I-IV,  40  cts.  Manual  to  Books  V-VIII, 
40  cts.  Two-Book  Course:  Book  A,  per  dozen,  $1.20;  Book  B,  per  dozen,  $1.75; 
Manual,  40  cts. 

Thompson's  .Esthetic  Series  of  Drawing.   This  series  includes  the  study  of 

Historical  Ornament  and  Decorative  Design.  Book  I  treats  of  Egyptian  art;  Book  II, 
Greek  ;  Book  III,  Roman  ;  Book  IV,  Byzantine  ;  Book  V,  Moorish ;  Book  VI,  Gothic. 
Per  dozen,  $1.50.    Manual,  60  cents. 

Thompson's  Educational  and  Industrial  Drawing. 

Primary  Free-Hand  Series  (Nos.  1-4).     Each  No.,  per  doz.,  $1.00.     Manual,  40  cts. 

Advanced  Free-Hand  Series  (Nos.  5-8.)     Each  No.,  per  doz.,  ^1.50. 

Model  and  Object  Series  (Nos.  1-3).     Each  No.,  per  doz.,  $1.75.     Manual,  33  cts. 

Mechanical  Series  (Nos.  1-6).     Each  No.,  per  doz.,  $2.00.     Manual,  75  cts. 
Thompson's  Manual  Training  No.  l.      Clay  modeling,  stick  laying,  paper  folding, 

color  and  construction  of  geometrical  solids.     lUus,     66  pp.     25  cts. 
Thompson's     Manual    Training    No.  2.      Mechanical  drawing,   clay    modelling, 

color,  wood  carving,     lUus.    70  pp.    25  cts. 
Thompson's  Drawing  Tablets.      Four  Tablets,  with  drawing  exercises  and  practice 

paper,  for  use  in  the  earlier  grades.     Each  No.,  per  doz.,  $1.20. 
Drawing  Models.      Individual  sets  and  class  sets  of  models  are  made  to  accompany' 

several  of  the  different  series  in  the  Thompson  Drawing  Courses.    Descriptive  circulars 

free  on  request. 

Anthony's  Mechanical  Drawing.     98  pages  of  text,  and  32  folding  plates.   |i.so. 
Anthony's  Machine  Drawing.     65  pages  of  text,  and  iS  folding  plates.   ;^i.so. 
Anthony's  Essentials  of  Gearing.     84  pages  of  text,  and  15  folding  plates,  ^1.50. 
Daniels's  Freehand  Lettering.     34  pages  of  text,  and  13  folding  plates.    75  cts. 

Johnson's  Lessons  in  Needlework.  Gives,  with  illustrations,  full  directions  for 
work  during  six  grades.     117  pages.     Square  8 vo.     Cloth,  $1.00.     Boards,  60  cts. 

Lunt's  Brushwork  for  Kindergarten  and  Primary  Schools.     Eighteen  lesson 

cards  in  colors,  with  teacher's  pamphlet,  in  envelope.    25  cts. 

Seidel'S  Industrial  Instruction  (Smith),  a  refutation  of  all  objections  raised  against 
industrial  instruction.     170  pages.    90  cents. 

Waldo's  Descriptive  Geometry.  A  large  number  of  problems  systematically  ar- 
ranged, with  suggestions.    85  pages.    80  cents. 

Whitaker's  How  to  use  Woodworking  Tools.    Lessons  in   the  uses  of  the 

hammer,  knife,  plane,  rule,  square,  gauge,  chisel,  saw  and  auger.     104  pagw.    60  cents. 

Woodward's  Manual  Training  School,     its  aims,  methods  and  results:  with 

detailed  courses  of  instruction  in  shop-work.     Illustrated.     374  pages.     Octavo.    $2.00. 
Setit  postpaid  by  mail  on  receipt  of  price. 

D.C.  HEATH  h  CO.,  Publishers,  Boston,  NewYork,  Chicago 


14  DAY  USE 

RETURN  TO  DESK  FROM  WHICH  BORROWED 

LOAN  DEPT. 

This  book  is  due  on  the  last  date  stamped  below,  or 

on  the  date  to  which  renewed. 

Renewed  books  are  subject  to  immediate  recall. 


i<s^ 


tM'- 


RSC'D  LD 


NOV  19  lasi 


^ 


m 


'3^ 


RECn  r  n« 


MflR  T 1 1962- 


— i(jm  gp^'*'*' 


LD  2lA-50m-8,'61 
(C1795sl0)476B 


General  Library 

University  of  California 

Berkeley 


llj 


lyUlH^^-^ 


^.d  'V>M^ 


V'vWiaiv'^, 


y 


918232 


^Af-rr 


THE  UNIVERSITY  OF  CALIFORNIA  LIBRARY 


->^Ji,^ 


